Infinite sequence of distinct spaces, all with same homology

This doesn't directly address you question, but since you said you were interested, here you go:

There is an infinite family of simply connected closed manifolds having isomorphic homotopy, homology, and cohomology groups in all dimensions which are distinguished by their cohomology ring. The spaces are all $S^2$ bundles over $\mathbb{C}P^2\sharp \mathbb{C}P^2$.

To begin with, There is a $T^2$ action on $S^3\times S^3$ given by $(z,w)\ast(p,q) = (zwp, z^2 w q)$, where we think of $p,q$ as unit quaternions and $z$ and $w$ as unit complex numbers. This action is free and the quotient space $(S^3\times S^3)/T^2$ is diffeomorphic to $\mathbb{C}P^2\sharp \mathbb{C}P^2$, as shown in Totaro's paper "Cheeger manifolds and the classification of biquotients"

Now, consider the $T^2$ action on $S^2$ given by $(z,w) p = z^m w^n \,p\, \overline{z}^m \overline{w}^n$, where we are thinking of $S^2$ as the unit length purely imaginary quaternions.

Then we have a diagonal action of $T^2$ on $S^2\times (S^3\times S^3)$. Projection onto the second factor gives the quotient space $M_{m,n} = [S^2\times (S^3\times S^3)]/T^2$ the structure of a bundle over $\mathbb{C}P^2\sharp \mathbb{C}P^2$ with fiber $S^2$. Using this and the Gysin sequence, one can easily show the homology and cohomology groups of $M_{m,n}$ are isomorphic to those of $S^2\times S^2\times S^2$, independent of $m$ and $n$.

We also have a principal bundle $T^2\rightarrow S^2\times(S^3\times S^3)\rightarrow M_{m,n}$. Using the LES in homotopy groups associated to this, it follows that $\pi_k(M_{m,n}) \cong \pi_k(S^2\times S^2\times S^2)$ for all $k$.

Finally, in my thesis, I computed the ring structure. The cohomology ring of $M_{m,n}$ is given by $\mathbb{Z}[u,v,w]/I$ where $I$ is the ideal generated by $u^2 + 2uv$, $uv + v^2$ and $muw + nvw + w^2$.

Then one can show that the number $m^2 + (m-n)^2$ is an invariant of the cohomology ring. In fact, I proved that $H^\ast(M_{m,n})\cong H^\ast(M_{m',n'})$ iff $m^2 + (m-n)^2 = m'^2 + (m'-n')^2$ and both $m-m'$ and $n-n'$ are even.