Seems to me like it is. There are only finitely many distinct powers of $x$ modulo $p$, by Fermat's Little Theorem (they are $\{1, x, x^2, ..., x^{p-2}\}$), and the coefficient that I choose for each of these powers can only be taken from $\{0,1,2,..., p-1\}$. So essentially I'm choosing amongst $p$ things $p-1$ many times, resulting in at most $p^{p-1}$ distinct polynomials.
Yet an assignment claims that $\mathbb{Z}_p[x]$ is infinite.
Here's the flaw in your reasoning: when you consider $x$, the variable of your polynomial, $x$ is not taken to be a number in $\mathbb Z_p$. In fact, $x$ is not really a number of any sort, and the polynomial is not meant to be interpreted as a function, but rather a sort of "algebraic object." Think of it as a tuple with a number of entries equal to the degree of the polynomial plus one.
You're thinking of $\mathbb Z_p[x]$ as the set of polynomial functions $f:\mathbb Z_p\mapsto \mathbb Z_p$. But the polynomials are not defined this way. They are defined only by their coefficients.
For example, $$x^{p+1}+1= x+1$$ if $x\in\mathbb Z_p$, but $$x^{p+1}+1\ne x+1$$ if the two above polynomials are regarded as elements of $\mathbb Z_p[x]$.
One last clarification: $$\mathbb Z_p[x]=\{a_0+a_1 x+...+a_n x^{n}: a_i\in\mathbb Z_p, n\in\mathbb N\}$$ However, you might have thought that $$\mathbb Z_p[x]=\{a_0+a_1 x+...+a_n x^{n}: a_i,x\in\mathbb Z_p, n\in\mathbb N\}$$
$\Bbb Z_p[x]$ is indeed infinite, since it contains polynomials, with coefficients in $\Bbb Z_p$, of arbitrary high degree, e.g. $x^n \in \Bbb Z_p[x]$, where $n \in \Bbb N$; also, $x^n \ne x^m$ if $m \ne n$.
Fermat's Little Theorem, that $a^p = a$ for $a \in \Bbb Z_p$, does not apply to the indeterminate $x \in \Bbb Z_p[x]$. $x \in \Bbb Z_p[x]$ and $a \in \Bbb Z_p$ are definitely "birds of a different feather" in this regard.
So though $\Bbb Z_p$ is finite, $\Bbb Z_p[x]$ is not.
If $p=3$, it's not true that $x=x^4$ in this ring $\mathbb Z_3[x]$. Only the coefficients are mod $p$. While $x^{p-1}$ might be the same as $1$ as a function, at least on $\mathbb Z_p^*$, it's not the same as a symbol which represents multiplying an indeterminate $x$ by itself $p-1$ times.
You might have $x^2\equiv x \bmod 2$ for the two elements $0$ and $1$. So this is within $\mathbb Z_2$.
Now note that $x^2+x+1$ has no roots in $\mathbb Z_2$, so let's invent one and call it $\alpha$, and there will be another root $\alpha+1$ because the sum of the roots is then $\alpha+(\alpha+1)\equiv 1\equiv -1 \bmod 2$.
This is the same process by which we invent/construct/discover/adjoin a root of the real polynomial $x^2+1$ and call it $i$.
We then find that $\alpha^2=\alpha+1$ (modulo $2$) so that the two polynomials $x^2$ and $x$ are no longer the same function when we add the element $\alpha$ to the mix, even if we are still working modulo $2$. This is a small example of something which becomes rather common when you are trying to analyse polynomials and their roots.
So no, the $x$ and $x^2$ are not formally the same modulo $2$, as other have said, but importantly they behave differently in important algebraic contexts and it matters that we should distinguish them.
$x^p-x$ is a degree $p$ polynomial. Whereas for $a\in\mathbb Z_p$ we have Fermat's little theorem, $x^p=x$ is not true in general.
In fact (by the factor theorem), $p(x)=x^p-x$ has at most $p$ roots.
So $n\not=m\implies x^n\not=x^m$.
$\mathbb Z_p[x]$ is in fact an infinite dimensional vector space over $\mathbb Z_p$, with basis $\{1,x,x^2,\dots\}$.
Hence it is an infinite set.
I think that you'll get a clearer view if you come back to the definition of a ring structure (commutative ring for simplification). The definitions being granted, let $A$ be a subring of a ring $B$. For $x\in B, A[x]$ is defined as being the smallest subring of $B$ containing $x$. You can immediately check that $A[x]$ consists of all the polynomial expressions $y=a_0 = a_1 x+...+a_nx^n$, with $a_i \in A$ and variable degree $n$. Your ring $\mathbf F_p [x]$ is an example, but a very restrictive one because you take $x \in \mathbf F_p$, so that $\mathbf F_p [x]=\mathbf F_p$ .
The ring of polynomials $A[X]$ (remark the change of notations, $X$ instead of $x$) is a different matter. In most textbooks (such as S. Lang's "Algebra", chap. V, §2), it is presented abstractly as a "universal object", which roughly means, without too much formalism, that any ring $A[x]$ as above is a quotient-ring of $A[X]$ via a (unique) surjective ring-homomorphism which sends $X$ to $x$. Concretely, a "prototype" of $A[X]$ will be the set of all sequences $\alpha:=(a_0 ,..., a_n,...)$ s.t. the $a_i\in A$ are almost all zero (i.e. all of them, except perhaps a finite number, are zero), endowed with a ring structure by putting $\alpha + \beta = (..., a_n +b_n,...)$ and $\alpha . \beta = (...,c_n ,...)$ s.t. $c_n=a_0b_n + a_1b_{n-1} +...+ a_nb_0$. The "indeterminate" $X$ will just be the sequence (0, 1, 0,...,0,...).
Coming back to your example, you have a surjective ring-homomorphism $\phi:\mathbf F_p[X] \to \mathbf F_p[x]$. If $\phi$ is injective, $x$ is transcendental over $\mathbf F_p$ and $\mathbf F_p[x] \cong \mathbf F_p[X]$ is infinite. If $\phi$ is not injective, $x$ is algebraic over $\mathbf F_p$ and $\mathbf F_p[x] \cong \mathbf F_p[X]/ker \phi$ is a finite field (why ?) ./.
