at.algebraic topology - Are there two non-homotopy equivalent spaces with equal homotopy groups?

The simplest examples I know are the $3$-dimensional lens spaces $L(p,q)$. They display many oddities.

Consider the lens spaces $L(p,q_0)$ and $L(p,q_1)$, $gcd(p,q_0)=gcd(p,q_1)=1$. Their fundamental groups are isomorphic to the cyclic group $newcommand{bZ}{mathbb{Z}}$ $bZ/pbZ$. Since both these lens spaces have the same universal cover $S^3$, their higher homotopy groups are also isomorphic.

A theorem of Franz-Rueff-Whitehead (see Theorem 2.60 of these notes) shows that $L(p,q_0)$ and $L(p,q_1)$ are homotopy equivalent if and only

$$q_1equiv pm ell^2 q_0bmod p, $$

for some $ellinbZ$. This reduces the problem to a number theoretic one. For example, $L(5,1)$ is not homotopy equivalent to $L(5,2)$ since $pm 2$ is not a quadratic residue mod $5$.

On the other hand, the lens spaces $L(7,1)$ and $L(7,2)$ are homotopy equivalent, but they are not homeomorphic.