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.
An easy example is the torus $S^1 \times S^1$, which has the same homology as but different cohomology ring than the wedge $S^1 \vee S^1 \vee S^2$ (which has no nontrivial cup products).
A more interesting question is whether there are examples which are both closed manifolds. There might be 3-manifold examples but I don't know how to construct them off the top of my head.
For 4-manifolds we can construct examples by finding simply connected closed 4-manifolds with the same middle Betti number $b_2$ but such that the absolute value of the signature is different, which implies that the cohomology rings are not isomorphic. I think we can take $\mathbb{CP}^2 \# \mathbb{CP}^2$ and $\mathbb{CP}^1 \times \mathbb{CP}^1$; these both satisfy $b_2 = 2$ but the first one has signature $2$ and the second has signature $0$ (although you don't need to know what signature is to compute that the cohomology rings aren't isomorphic). See this blog post for more background.
Best Answer
Standard example is $\mathbb RP^2\times S^3$ and $\mathbb RP^3\times S^2$ (they have same homotopy groups since they both have $\pi_1=\mathbb Z/2$ and the universal cover is in both cases $S^2\times S^3$).