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 and the map $g$ is birational.  

**EDIT**:  My argument that $f^{-1}(Z)$ was reduced was junk.  I shouldn't have tried to do math while on the run.  But as long as $f^{-1}(Z)$ is reduced and $g$ is birational, then I think things are ok.