Let's suppose $M$ is a nonzero projective $\Z$-module. Can it be finite? Nope. I'm sure there are plenty ways to prove it, but one way is to observe that a projective $\Z$-module is free, and hence if $M$ is nonzero it must have at least one copy of $\Z$. So, $M$ is infinite.

What's the analogue for the ring $k[x]$ where $k$ is a field? If $M$ is a nonzero projective $k[x]$-module, can it be finite? It certainly can't if $k$ is infinite dimensional, since any nonzero $k[x]$-module (whether projective or not) is also a $k$-vector space. What about if $k$ is finite?

To answer this question, let's ask a better question: if $M$ is a nonzero projective $k[x]$-module, can it be *finite-dimensional* as a $k$-vector space? Ah, excellent question. To answer this, let's suppose $M$ is a $k[x]$-module that is finite-dimensional as a $k$-vector space. Then, for some $n$ there exists an exact sequence

$$0\to A\to k[x]^n\to M\to 0$$ of $k[x]$-modules. If $M$ is projective, it gives an isomorphism $A\oplus M\to k[x]^n$. This shows that $M$ is isomorphic to some $k[x]$-submodule of $k[x]^n$. But, any nonzero $k[x]$-submodule of $k[x]^n$ is infinite-dimensional as a $k$-vector space. So, if $M$ is indeed a projective $k[x]$-module, it must be infinite-dimensional.

Of course, you can always just use the fundamental structure theorem for principal ideal domains. If $M$ is a projective and finitely-generated $k[x]$-module then it must be free and hence in particular infinite dimensional as a $k$-vector space. However, you don't have to use the structure theorem.

But let's see a concrete example! Take any field $k$. (Here you have to wonder whether I mean *nonzero* field. Most people do.) Consider the $k[x]$ module $k$ where $x$ acts by multiplication by an element $\alpha\in k$. We have a surjection $k[x]\to k$ defined by $1\mapsto 1$. This surjection is of course well-defined because $k[x]$ is a free $k[x]$-module with $1$ as a generator.

If we apply the functor ${\rm Hom}_{k[x]}(k,-)$ to this surjection, then we get a map

$${\rm Hom}_{k[x]}(k,k[x])\rightarrow{\rm Hom}_{k[x]}(k,k).$$ I claim that this map is not surjective, which if true would show that $k$ cannot be projective. To prove this claim, observe first that any $k$-linear map $k\to k$ commutes with multiplication by $\alpha$, and hence is also a $k[x]$-module homomorphism. On the other hand, suppose that $\psi:k\to k[x]$ is a $k[x]$-module homomorphism. Then for any $\beta\in k$, we have $\psi(x\beta) = x\psi(\beta)$. The left-hand side of this equation is the constant $\alpha\beta$, whereas the right-hand side is a polynomial of degree at least one **unless** $\psi(\beta) = 0$. Therefore, the only possibility is that $\psi(\beta) = 0$. So, the map

$${\rm Hom}_{k[x]}(k,k[x])\rightarrow{\rm Hom}_{k[x]}(k,k)$$ is the map $0\to k.$ Certainly not surjective! Therefore, $k$ cannot be projective.

So, in closing, it actually doesn't matter whether $k$ is finite. A nonzero projective $k[x]$-module not only can't be finite, it can't be finite-dimensional, and this is true for finite and infinite fields.