The answer to this question depends very much on how you define a triangle, as there is not really a single consistent definition which is used throughout mathematics. Quoting MathWorld:
A polygon can be defined (as illustrated above) as a geometric object "consisting of a number of points (called vertices) and an equal number of line segments (called sides), namely a cyclically ordered set of points in a plane, with no three successive points collinear, together with the line segments joining consecutive pairs of the points. In other words, a polygon is closed broken line lying in a plane" (Coxeter and Greitzer 1967, p. 51).
There is unfortunately substantial disagreement over the definition of a polygon. Other sources commonly define a polygon (in the sense illustrated above) as a "closed plane figure with straight edges" (Gellert et al. 1989, p. 162), "a closed plane figure bounded by straight line segments as its sides" (Bronshtein et al. 2003, p. 137), or "a closed plane figure bounded by three or more line segments that terminate in pairs at the same number of vertices, and do not intersect other than at their vertices" (Borowski and Borwein 2005, p. 573). These definitions all imply that a polygon is a set of line segments plus the region they enclose, though they never define precisely what is meant by "closed plane figure" and universally depict polygons as a closed broken black lines with no shading of the interiors.
There are basically two definitions which might be relevant:
Definition A: A triangle is a plane figure bounded by three line segments which terminate at three vertices.
or
Definition B: A triangle is a geometric object consisting of three points (vertices) and three line segments (sides).
Assuming Defintion A:
If we assume Definition A, then the answer is that there are no triangles contained in the Sierpinski gasket. The simplest argument which show this is, I think, a topological argument. A triangle has nonempty interior (that is, in any triangle we can find at least one point $x$ and some positive number $r > 0$ such that the ball of radius $r$ centered at $x$ is entirely contained in the triangle). On the other hand, the Sierpinski gasket has empty interior. As the Sierpinki gasket has empty interior, it cannot contain any triangles.
Assuming Defintion B:
If we assume Definition B, then the answer is more interesting. For specificity of notation, let $\mathscr{T} \subseteq \mathscr{P}(\mathcal{S})$ denote the set of triangles contained in the Sierpinski gasket.
One construction of the Sierpinski gasket is as the attractor of an iterated function system. Specifically, define the contraction maps
\begin{align}
\varphi_1(x,y) &:= \tfrac{1}{2}(x,y) \\
\varphi_2(x,y) &:= \tfrac{1}{2}(x,y) + (\tfrac{1}{2},0) \\
\varphi_3(x,y) &:= \tfrac{1}{2}(x,y) + (\tfrac{1}{4},\tfrac{\sqrt{3}}{4}).
\end{align}
The Sierpinski gasket $\mathcal{S}$ is the unique nonempty compact set satisfying the relation
$$ \mathcal{S} = \bigcup_{j=1}^{3} \varphi_j(\mathcal{S}). $$
This iterated function system gives us means for assigning an "address" to many (all?) of the triangular subsets of the gasket. In particular, if a triangle $T$ is contained in the Sierpinski gasket, then so too is its image under each of the maps in the iterated function system. So we can start by identifying an "ur-triangle", then watching as that triangle is mapped forward by the iterated function system.
Let $T$ be the triangle with vertices
$$ \left(0,0\right), \qquad \left( 1, 0 \right), \qquad\text{and}\qquad \left( \frac{1}{2}, \frac{\sqrt{3}}{2}\right). $$
Note that $T \in \mathscr{T}$, i.e. this triangle is contained in the Sierpinski gasket (indeed, this is the convex hull of the gasket with respect to this construction, and is therefore the largest triangle contained therein).
After applying the iterated function system to $T$, we obtain the following:
In this image, $T_j = \varphi_j(T)$, and $G$ is the triangle bounded by the $T_j$. (I have labeled it $G$ for "gap": one common construction of the Sierpinski gasket is to remove solid triangles at each stage of the construction leaving gaps; $G$ is the first such removed triangle). Observe that $T_j, G \in \mathscr{T}$.
Applying the iterated function system again, we get
For each $j$, the set $G_j$ is $\varphi_j(G)$, and the sets $T_{jk}$ are the sets $\varphi_{k} \circ \varphi_{j}(T)$. Note that (1) all of the sets $G_j$ and $T_{jk}$ are triangles contained in the Sierpinski gasket, and (2) we have not relabeled the triangle $G$, as it has already been counted (in the previous stage of the construction). Again, $T_{jk},G \in \mathscr{T}$.
We can continue this process: for any finite sequence $\alpha$ consisting of the digits $1$, $2$, and $3$ (that is, for any $\alpha = (\alpha_1, \alpha_2, \dotsc, \alpha_m) \in \bigcup_{n=1}^{\infty} \{1,2,3\}^n$) let
$$ T_{\alpha} = \varphi_{\alpha_m} \circ \dotsb \circ \varphi_{\alpha_2} \circ \varphi_{\alpha_1} (T)
\qquad\text{and}\qquad
G_{\alpha} = \varphi_{\alpha_m} \circ \dotsb \circ \varphi_{\alpha_2} \circ \varphi_{\alpha_1} (G). $$
For any $\alpha$, we have $T_{\alpha},G_{\alpha} \in \mathscr{T}$. Therefore
$$ \{T, G\} \cup \left(\bigcup_{n=1}^{\infty} \bigcup_{\alpha\in \{1,2,3\}^n} T_{\alpha} \right) \cup \left(\bigcup_{n=1}^{\infty} \bigcup_{\alpha\in \{1,2,3\}^n} G_{\alpha} \right) \subseteq \mathscr{T}. $$
This set of triangles is the countable union of finite sets, therefore countable. Thus $\mathscr{T}$ is at least countable.
To fully answer the question, it remains to show that the above set is exactly $\mathscr{T}$, which would tell us that the set of triangles contained in $\mathcal{S}$ is countable. This can be done by observing that we have already counted all of the triangles with sides parallel to the sides of $T$ (this should be checked), and that any segment that is not parallel to the sides of $T$ must eventually intersect one of the removed triangles in the construction (this should also be checked). Hence such a segment is not entirely contained in the gasket, and therefore cannot be the side of a triangle contained in the gasket.
From this, it follows that the Sierpinski gasket contains countably many triangular subsets (all of which are equilateral, even).
