Same homology but different fundamental group

Let $M$ and $N$ be connected, closed, orientable $3$-manifolds. Then I'm able to prove the following statement :

If $\pi_1(M)\simeq\pi_1(N)$ then $H_i(M)\simeq H_i(N)$ for all $i$.

But I wonder if the converse is also true.

If $H_i(M)\simeq H_i(N)$ for all $i$ then $\pi_1(M)\simeq\pi_1(N)$.

I believe the statement is false and tried to find counterexample. Since the abelianization of fundamental group is the first homology group, I think the possible counterexample would contain $\Bbb CP^n$ or $\Bbb RP^n$ with some restriction on $n$ ($M$ and $N$ should be orientable). But it seems such counterexamples is not very easy to find so now I'm guessing the statement is in fact true. Could you help?

The Poincare homology sphere has the same homology as the $3$ sphere but a nontrivial fundamental group.