Let $R$ be a commutative ring such that for some integer $n \ge 1$, the ring $\mathbb Z^n$ injects into $R$ and also there is a split surjective ring homomorphism $R \to \mathbb Z^n$ whose kernel is the nilradical of $R$. So in particular, $R \cong \mathrm{Nil}(R) \oplus \mathbb Z^n$ as abelian groups.
My question is:
If $R \otimes_{\mathbb Z} \mathbb Q$ is a finite dimensional $\mathbb Q$-vector space, then is $R$ a finitely generated $\mathbb Z$-module ?
Note: If you are looking for a concrete example of a ring $R$ with the injective and split surjective conditions as mentioned ... the $K_0$ of any Commutative Noetherian ring is such an example ...
No. Indeed, for any abelian group $N$, you can make $R=N\oplus\mathbb{Z}$ into a ring with multiplication $(x,a)\cdot(y,b)=(bx+ay,ab)$ (in other words, you take formal sums of integers and elements of $N$ with $N^2=0$). This ring satisfies your hypotheses, and $R\otimes\mathbb{Q}$ is finite-dimensional as long as $N\otimes\mathbb{Q}$ is finite-dimensional. So you could take $N$ to be any infinite torsion group to get a counterexample.