A standard example of two CW complexes which have isomorphic homotopy groups but are not homotopy equivalent is $ RP^2 \times S^3$ and $RP^3 \times S^2$.
The easiest way to see that they are not homotopy equivalent is by looking at their homology (i.e. ala the Kunneth formula).
What is an example of two CW complexes with isomorphic homotopy groups and homology, yet not homotopy equivalent?
It seems like an obvious place to look would be to have non-isomorphic cohomology (as rings). If there is such an example, then the next question would be for an example of two non-homotopy equivalent CW complexes with isomorphic homotopy groups, and isomorphic homology (as graded groups) and cohomology (as rings). Do such examples exist? I suspect they do, but I don't know any.
Update: there are good answers below to the first question, with links. But it seems to still beg the question about cohomology: does anyone know an example of two non-homotopy equivalent CW complexes with isomorphic homotopy groups, homology groups, and cohomology rings?
Best Answer
Sorry to be late to the party, but here's an example of 2 compact simply connected manifolds which have the same homotopy groups, same homology groups, same cohomology ring, and yet are not homotopy equivalent. The examples are motivated by Grigory M's examples:$S^2\times S^2$ and $\mathbb{C}P^2\#\overline{\mathbb{C}P^2}$. His examples are both $S^2$ bundles over $S^2$.
If we extend this further, it turns out there are precisely two $S^3$ bundles over $S^2$. Of course, one is the product $S^3\times S^2$, while another doesn't have a more common name, so I'll just denote it $S^3\hat{\times} S^2$.
Both of these spaces are diffeomorphic to quotients of free linear $S^1$ actions on $S^3\times S^3$. Letting $X$ denote either bundle, we have a long exact sequence of homotopy groups $$...\pi_k(S^1)\rightarrow \pi_k(S^3\times S^3)\rightarrow \pi_k(X)\rightarrow \pi_{k-1}(S^1)\rightarrow ...$$
which can be used to show that $\pi_k(X) \cong \pi_k(S^3\times S^3)$ for $k\geq 2$ and $\pi_1(X) = \{e\}$ and $\pi_2(X) \cong \mathbb{Z}$.
The Hurewicz theorem together with Universal coefficients theorem implies $H^1(X) = 0$ and $H^2(X) \cong \mathbb{Z}$. Poincare duality then forces the rest of the cohomology rings to agree.
Finally, to see $S^3\times S^2$ and $S^3\hat{\times} S^2$ are different, one computes the Stiefel-Whitney classes of their tangent bundles. It turns out $w_2(S^3\times S^2) = 0$ while $w_2(S^3\hat{\times}S^2)\neq 0$. (And all other Stiefel-Whitney classes are $0$ for both spaces). Since the Stiefel-Whitney classes can be defined in terms of Steenrod powers, they are homotopy invariants, so $S^2\times S^3$ and $S^3\hat{\times}S^2$ are not homotopy equivalent.