Question: Let $X$ and $Y$ be path-connected spaces that admit a contractible universal cover, with $\pi_1(X) \cong \pi_1(Y)$. Is $X$ homotopy equivalent to $Y$?
Comments: $X$ and $Y$ are both $K(\pi_1(X),1)$s. In particular, this implies that every homomorphism $\varphi: \pi_1(X) \rightarrow \pi_1(Y)$ (e.g., the isomorphism) is induced by a map $f: X \rightarrow Y$. If $X$ and $Y$ are both CW-complexes, Whitehead's theorem says that $f$ is a homotopy equivalence. (In general, by definition, $f$ is a weak homotopy equivalence.) So a counterexample requires that at least one of $X$ and $Y$ fails to have homotopy type a CW-complex.
If one removes the requirement that $X$ and $Y$ have contractible universal cover, in particular if one even relaxes it to $X$ and $Y$ have weakly contractible universal cover, the double comb space is a simply-connected counterexample, as it is not contractible. (A proof that it is not contractible can be found here.)