Let $M$ be a simply connected, (finite dimensional) smooth manifold. Is it possible that $M$ is homotopy equivalent to $M\times M,$ without $M$ being contractible? This would imply $\pi_n(M)\times\pi_n(M)\cong \pi_n(M)$ for all $n\in\mathbb{N}.$ I know there are groups which satisfy $G\times G\cong G,$ but this is a very strong condition, and this condition still seems much weaker than the condition in question.

According to When is $G$ isomorphic to $G \times G$?, if even one nontrivial homotopy group is finitely generated, this is impossible.

(I asked this on stackexchange and didn't get any responses.)