My understanding is Fermat Last Theorem
$$\not\exists x, y, z \in \mathbb{Z}_{+} : x^p + y^p = z^p\quad\text{ for } p > 2$$ is plausible because integers of $p^{th}$ power thin out too quickly as $x$ getting big (at least for $p > 3$ ).
For any integer $N$, the number of $p^{th}$ power smaller than $N$ is of the order
of $N^{\frac{1}{p}}$. If we randomly pick a integer near an number $N$, the probability that we get a $p^{th}$ power is around $\frac{1}{p} N^{\frac{1}{p}-1}$.
Heuristically, the probability that we get a sum of two $p^{th}$ power should be something like
$$\frac{1}{p^2} \sum_{n=1}^{N-1} n^{\frac{1}{p}-1} (N-n)^{\frac{1}{p}-1}
\sim \frac{1}{p^2} N^{\frac{2}{p}-1} \int_0^1 x^{\frac{1}{p}-1}(1-x)^{\frac{1}{p}-1} dx
= \frac{\Gamma\left(\frac{1}{p}\right)^2}{p^2\Gamma\left(\frac{2}{p}\right)} N^{\frac{2}{p}-1}
$$
Replace $N$ by the list of $p^{th}$ power and sum over it, one will expect
the total number of primitive solutions $\mathcal{N}(p)$ for $x^p + y^p = z^p$ is of the order
$$\mathcal{N}(p) \sim \mathcal{N}_{est}(p) := \frac{\Gamma\left(\frac{1}{p}\right)^2}{p^2\Gamma\left(\frac{2}{p}\right)} \sum_{n=2}^\infty n^{2-p}
= \frac{\Gamma\left(\frac{1}{p}\right)^2(\zeta(p-2)-1)}{p^2\Gamma\left(\frac{2}{p}\right)}
\tag{*1}$$
Since the sum $\sum\limits_{n=2}^\infty n^{2-p}$ converges for $p > 3$,
we expect $\mathcal{N}(p)$ to be finite. When we plug in some actual numbers, we find
$$\begin{align}
\mathcal{N}_{est}(4) &\sim 0.29893898046807\\
\mathcal{N}_{est}(5) &\sim 0.076793757848265\\
\mathcal{N}_{est}(6) &\sim 0.026448003085251\\
&\;\vdots
\end{align}
$$
They are so small and we don't expect $\mathcal{N}(p)$ to be non-zero at all.
Of course, this argument only works for $p > 3$. I have no idea how to argue for the case $p = 3$.
