A consequence of Nullstellensatz is that if $K = \overline{K} $ and $V(I)$ has finitely many points then $\dim(A/I) = 0$.
Let $\alpha_i \in V(I)$. we denote with $m_{\alpha_i}$ the maximal ideal $I(\{\alpha_i\})$ $$I(V(I)) = \sqrt{I} = \bigcap_{i=1}^{|V(I)|} I\{\alpha_i\} =\bigcap_{i=1}^{|V(I)|} m_{\alpha_i} \implies A/I \cong \prod A/m_{\alpha_i} \cong K^s$$ What can we say about the krull dimension when the field is not algebraically closed? I think that the equality does not hold but i cant find a counterexample for $\dim(A/I) \neq 0$.
Take $K=\mathbb{R}$, $A=\mathbb{R}[x,y]$ and $I=(x^2+y^2+1)$. Then $V(I)$ is empty but $\dim(A/I)=1$.
Indeed, note that $x^2+y^2+1$ is irreducible: as polynomial of $\mathbb{R}[x][y]$ it is primitive, and hence by Gauss' Lemma it suffices to see irreducibility over $\mathbb{R}(x)[y]$. A quadratic polynomial over a field is irreducible if and only if it has no roots, so suppose by contradiction that we have $x^2+y^2+1=(y-p(x))(y+p(x))$ for some $p(x)\in\mathbb{R}(x)$. Then we must have $x^2+1=-(p(x))^2$, which is impossible (we can evaluate $p$ at a point where its denominator doesn't vanish, and then LHS is positive while RHS is negative).
Thus $A/I$ is a domain. As $I$ is strictly contained in the maximal ideal $(x,y)$ of $A$, the domain $A/I$ admits a non-trivial maximal ideal. Hence $\dim(A/I)\geq 1$, which suffices for the purposes of the question.
With a bit of background it is in fact fairly straightforward to then also show that in fact equality holds.