Can't we base-change $f : X \to Y$ with $Z$ and obtain: $g : f^{-1}(Z) \to Z$?  This also is a universal homeomorphism by construction, right?  So now we have a universal homeomorphism to a regular scheme, but a regular scheme is *weakly normal*, see A. Andreotti and E. Bombieri, ``Sugli omeomorfismi delle varietà algebriche''.  

Therefore $g$ is an isomorphism at least as long as $f^{-1}(Z)$ is reduced.  But $f^{-1}(Z)$ has to be reduced, because if it isn't reduced this would imply inseparability of $f$ at some (possibly generic) point of $Z$, which contradicts the universal homeomorphism of $f$.

I have to run for a couple hours, but did I misunderstand the question?