All of these examples involve a bit of cleverness, so I thought I'd point out a more straightforward way to construct counterexamples. If X is any space, we can build a space X' = K(π₀X, 0) × K(π₁X, 1) × K(π₂X, 2) × ... which has the same homotopy groups as X, but which is usually not weakly equivalent to X. Pretty much any invariant of spaces other than the homotopy groups will distinguish them. For instance, if X = S², then H₃(X) = 0, but X' contains K(π₃S², 3) = K(Z, 3) as a retract, so H₃(X') contains H₃(K(Z, 3)) = Z as a retract and cannot be 0.

Of course, this doesn't produce geometrically nice spaces like smooth manifolds.

All of these examples involve a bit of cleverness, so I thought I'd point out a more straightforward way to construct counterexamples. If $X$ is any space, we can build a space $X' = K(\pi_0 X, 0) \times K(\pi_1 X, 1) \times K(\pi_2 X, 2) \times \dots$ which has the same homotopy groups as $X$, but which is usually not weakly equivalent to $X$. Pretty much any invariant of spaces other than the homotopy groups will distinguish them. For instance, if $X = S^2$, then $H_3(X) = 0$, but $X'$ contains $K(\pi_3 S^2, 3) = K(\mathbb{Z}, 3)$ as a retract, so $H_3(X')$ contains $H_3(K(\mathbb{Z}, 3)) = \mathbb{Z}$ as a retract and cannot be $0$.

Of course, this doesn't produce geometrically nice spaces like smooth manifolds.