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.