$\operatorname{Spec}(k[x])$ has infinite points. 
Let $k$ be a field. I have to prove that $\operatorname{Spec}(k[x])$ has infinite points. 

If $k$ is infinite it is obvious: in fact there are infinite maximal ideals $(x-\alpha_i)$, with $a_i \in k$. But if $k$ is a finite field, how can I prove my claim? Is there a general argument like the proof of the infinitude of primes of $\mathbb{Z}$?
 A: You can reduce the finite case to the infinite case.
Consider the morphism  $\mathrm{Spec}(\overline{k}[x]) \to \mathrm{Spec}(k[x])$. The fiber of $(0)$ is $\mathrm{Spec}(\overline{k}(x))$, just one point, and the fiber of $(f)$ with $f \in k[x]$ irreducible is $\mathrm{Spec}(\overline{k}[x]/(f))$, which is a finite set. It follows that if $\mathrm{Spec}(\overline{k}[x])$ is infinite, the same is true for $\mathrm{Spec}(k[x])$, and they have the same cardinality.
Actually this "geometric" proof can be made more "algebraic": There are infinitely many elements in $\overline{k}$. Since every polynomial over $k$ has only finitely many roots in $\overline{k}$, there are also infinitely many elements in $\overline{k}$ which are not conjugate to each other (pairwise). Hence, their minimal polynomials generate pairwise distinct maximal ideals in $k[x]$.
A: The maximal ideals in $k[x]$ are principal, generated by irreducible polynomials. Then you have to prove that there are infinitely many (non-associate) irreducible polynomials. In order to do this use the Euclid's trick for proving that there are infinitely many primes.
