I couldn't even find all the irreducible modules over $\mathbb{C}[x,y]$ yet.
I know that modules over $\mathbb{C}[x,y]$ is the same as $\mathbb{C}$-vector spaces $V$ equipped with two commuting operators $A, B \in End(V), \; [A,B]=0$.
I also can prove that any $\mathbb{C}[x,y]$-module $V$ contains finite dimensional $x$-stable subspace as well as $y$-stable subspace. It makes me think that all the irreducible $\mathbb{C}[x,y]$-modules are at least finite dimensional. Am I right? How do I find them?
The second part (finding all the completely reducible modules) I didn't actually try to solve yet because I obviously need to understand the first part to do that. It would be nice to have a hint for that.
Okay well a more general version of the "two commuting diagonalizable matrices are simultaneously diagonalizable" theorem is the theorem that any two commuting square matrices are simultaneously upper-triangularizable. It is essentially equivalent to the statement you are trying to prove though. Here is a sketch of how to prove it:
Suppose we have a finite dimensional $\mathbb{C}$ vector space with two commuting linear maps $X$ and $Y$. Since $\mathbb{C}$ is algebraically closed, there exists an eigenvector for $X$ with some eigenvalue $\lambda$. Since $X$ and $Y$ commute, $Y$ acts on the $\lambda$ eigenspace for $X$, so inside there we can again find an eigenvector but this time for $Y$.
Hence we have found a common eigenvector for $X$ and $Y$, if we quotient by this the subspace spanned by this we can repeat this process inductively and obtain the desired upper triangular form.