Well, for any ring epimorphism $f: R\to S$, the quotient $R/\mathrm{ker} f$ and $S$ have the same fraction field, so if you're working in a context where Krull dimension matches transcendence degree of the fraction field over a base field (finitely generated algebras over a field), then you're set.

First, pick a maximal chain of prime ideals in $S$ and mod out by the minimal one. Now $S$ is an integral domain of the same dimension. Similarly, you might as well assume $f$ is injective, since that can only decrease the Krull dimension of $R$.

So, now, we have a map, which must induce an isomorphism on fraction fields, and both algebras inject into their fraction fields. Now, take an ideal $I\subset S$ such that $I\cap R=0$, and let $s\neq 0$ be an element of $I$. Then $s=r'/r''$ for $r',r''\in R$. Thus $sr''\in R\cap I$, and we have arrived at a contradiction.