Dimensions of $Pol(\mathbb{Z}_3)$ (polynomial vector space) How many dimensions does $Pol (\mathbb{Z}_3)$ have, where $Pol (\mathbb{Z}_3)$ is a vector space of polynomial functions with one variable ($f: x \mapsto \sum^n_{k=0} \lambda_kx^k$).

My "guess" would be: infinite. Since $$\lambda_{n+1}x^{n+1} = \sum^n_{k=0} \lambda_kx^k$$ only holds for $\lambda_1 = \lambda_2 = \dots = \lambda_n = \lambda_{n+1} = 0$. So, the search for the biggest linear independent generator set never ends. So $dim(Pol (\mathbb{Z}_3)) = \infty$.
Let me denote by $\mathbb{Z}_3[X]$ the vector space of polynomials over $\mathbb{Z}_3$ in one indeterminate and by $\mathcal{Fun}(\mathbb{Z}_3)$ the vector space of functions from $\mathbb{Z}_3$ to itself. Any polynomial $p(X) = \sum_{i=0}^n p_iX^i$ in $\mathbb{Z}_3[X]$ induces an element of $\mathcal{Fun}(\mathbb{Z}_3)$ by evaluation as follows
$$\mathbb{Z}_3 \to \mathbb{Z}_3, \qquad z \mapsto \sum_{i=0}^{n}p_i\cdot z^{i}.$$
This defines a surjective morphism of vector spaces
$$\mathbb{Z}_3[X] \to \mathrm{Pol}(\mathbb{Z}_3) \tag{$\star$}$$
(in fact, this is the definition of $\mathrm{Pol}(\mathbb{Z}_3)$) which however is not injective. For example, the polynomial $X^3$ is sent to the function
$$[0]_3 \mapsto [0]_3, \qquad [1]_3\mapsto [1]_3, \qquad [2]_3\mapsto [8]_3 = [2]_3$$
which is the same as the image of the polynomial $X$ (in fact, by Fermat's Little Theorem).

By resorting to your argument, you proved that $\mathbb{Z}_3[X]$ is infinite dimensional. However, you may verify that any polynomial of the form $q(X)(X^3-X)$ is mapped to the zero function on $\mathbb{Z}_3$ by the assignment above. Conversely, if a polynomial $p(X)$ is mapped to the zero function then it admits $[0]_3,[1]_3,[2]_3$ as roots and hence, by the Factor Theorem,
$$p(X) = q(X)(X-[0]_3)(X-[1]_3)(X-[2]_3) = q(X)(X^3-X).$$
Summing up, the vector space of polynomial function on $\mathbb{Z}_3$ is $3$-dimensional.

As a bonus, you may check that $(\star)$ in fact exists for every field $K$ (and not only for $K=\mathbb{Z}_3$). However, it is bijective if and only if $K$ is infinite.
If $\mathbb{Z}_3$ means $\mathbb{Z}/3\mathbb{Z}$, then the dimension is not infinite. This is because you explicitly write "polynomial functions". A function is defined by its values on all inputs, and since $\mathbb{Z}/3\mathbb{Z}$ has only 3 elements, a function can only take 3 values. Therefore, there are exactly 9 functions on $\mathbb{Z}/3\mathbb{Z}$, so the dimension has to be finite. To find the dimension, try to look at the functions $1$, $x$, $x^2$ and so on.

If by $\mathbb{Z}_3$ you mean the 3-adic integers, then your argument works.