Constructing two non-isomorphic irreducible $\mathbb Q[x]$ modules with the underlying abelian group $\mathbb Q\times\mathbb Q$.

I'm trying a problem which asks me to construct two non-isomorphic irreducible $\mathbb Q[x]$ modules with the underlying abelian group $\mathbb Q\times\mathbb Q$. I've managed to prove (using isomorphism theorem) that irreducible modules are $\mathbb Q[x]/I$ with maximal ideals $I$. So I take the ideals $I_1=(x^2-2), I_2=(x^2-3)$. Then $\mathbb Q[x]/I_1$ and $\mathbb Q[x]/I_2$ are irreducible and also not isomorphic because any isomorphism will yield an isomorphism between $\mathbb Q(\sqrt 2)$ and $\mathbb Q(\sqrt 3)$.

My question is: Is the above argument ok? Also how do I rigorously argue for the part concerning the underlying groups? My modules are 'isomorphic' to $\mathbb Q(\sqrt 2)$ and $\mathbb Q(\sqrt 3)$, and intuitively the groups are $\mathbb Q\times\mathbb Q$ since any element in, say, $\mathbb Q(\sqrt 2)$, can be seen as an ordered pair $(a,b)$ or rationals. But how do I make this precise?

Your choice is perfectly fine. For the isomorphism of abelian groups, you know that the residue classes of $1,x$ form a $\mathbb Q$-basis of $\mathbb Q[x]/(x^2-2)$, i.e. we have an isomorphism $\mathbb Q[x]/(x^2-2) \cong \mathbb Q^2 = \mathbb Q \times \mathbb Q$ as $\mathbb Q$-vector spaces, in particular of abelian groups.
A $\mathbb Q[x]$-module structure of $\mathbb Q^2$ is the same as an endomorphism of $\mathbb Q^2$ i.e. 2x2-matrix. The module is irreducible iff the characteristic polynomial is irreducible. You picked the matrices $\begin{pmatrix}0&2\\1&0\end{pmatrix}$ and $\begin{pmatrix}0&3\\1&0\end{pmatrix}$.