Let $k$ be a non-algebraically closed field. Suppose $I\subset k[x_1,\dots, x_n]$ and consider $V_k(I)$(i.e. $k$ rational points of ideal $I$ in $k^n$). Denote $A(V(I))=k[x_1,\dots, x_n]/I$.
$\textbf{Q:}$ Is $V(I)$ finite set equivalent to $A(V(I))$ a finite dimensional vector space over $k$? My feeling is that $Spec(A(V(I))$'s closed points is finite. (Since $A(V(I))$ is finite dimensional over $k$, it is artinian. In particular, it is product of $k[x_1,\dots,x_n]/M_i$ with $M_i$ maximal ideals.) So it reduces to see the single $M_i$ case. So either $V(M_i)$ is empty or finite point sets.
This reduces the question to whether $V(M_i)$ is finite iff $k[x_1,\dots, x_n]/M_i$ is finite dimensional over $k$.
$\textbf{Refined Q:}$ Is $V(M_i)$ finite equivalent to $k[x_1,\dots, x_n]/M_i$ finite dimensional over $k$ with $M_i$ maximal?(It suffices to look geometric points of $V(M_i)$, but I am not totally convinced that $V(M_i)$ must be made of finite number points.)
It is known that if $C$ is a curve over $\mathbb{Q}$ of genus $g\geq 2$ then the rational points on $C$ form a finite set. But the quotient $\mathbb{Q}[x,y]/(f(x,y))$ should not form a finite dimensional vector space over $\mathbb{Q}$.