Let $\alpha,\beta\in\mathbb C$ be algebraic integers, so there exist monic $p,q\in\mathbb Z[x]$ such that $p(\alpha)=q(\beta)=0$. It follows that $\mathbb Z[\alpha,\beta]$ is finitely-generated as a $\mathbb Z$-module.
I want to show directly that any submodule of $\mathbb Z[\alpha,\beta]$ is finitely generated.
I'm aware that the result is true in general, since $\mathbb Z$ is a PID and all submodules of a finitely generated module over a PID are finitely generated. But I'm curious if there is a particular direct way in the above case, without going through the general argument.
[Converting my comment into an answer.]
When $\beta=0$ this amounts to proving that any submodule of $\mathbb{Z}^n$ is finitely generated (where $n$ is the degree of the minimal polynomial of $\alpha$). If you know that, it follows immediately that a submodule of a finitely generated $\mathbb{Z}$-module is finitely generated, by writing your finitely generated module as a quotient of $\mathbb{Z}^n$ for some $n$.
So, if you had a simple proof of your special case, you could very easily get a similarly simple proof that every submodule of a finitely generated $\mathbb{Z}$-module is finitely generated. As a result, I would not expect any easier proof to exist in your special case.