It is a well know fact that if $k \subset R$ is an extension of rings such that $R$ is a finite dimensional vector space over $k$, then every point of $Spec(R)$ is closed (i.e., equivalently every prime ideal of $R$ is maximal), and $Spec(R)$ is finite as a set. Here $Spec(R)$ = {prime ideals of $R$}.
If on the other hand we assume that $R$ is a ring and not a field, $Spec(R)$ consists of only closed points and $Spec(R)$ is finite as a set, then can we say anything about $dim_k(R)$, i.e., whether $R$ is a finite dimensional vector space over $k$? If not, what is a counterexample?
Also, what additional hypotheses might we need to make this statement work, if it doesn't?