For a pair of smooth, simply connected, compact, oriented 4-manifolds $X$ and $Y$,
Any isomorphism of the intersection lattices $H^2(X)\to H^2(Y)$ comes from an oriented homotopy equivalence $Y\to X$ (Milnor, 1958).
Any oriented homotopy equivalence is a tangential homotopy equivalence (Milnor, Hirzebruch-Hopf 1958).
Any oriented homotopy equivalence comes from an h-cobordism (Wall 1964).
Any oriented homotopy equivalence comes from a homeomorphism (Freedman).
It need not be the case that $X$ and $Y$ are diffeomorphic (Donaldson). Many examples are now known: e.g., Fintushel-Stern knot surgery on a K3 surface gives a family of exotic K3's parametrized by the Alexander polynomials of knots.
Here's a sketch of why homotopy equivalences preserve tangent bundles: $X$ and $Y$ have three characteristic classes: $w_2$, $p_1$ and $e$. However, $e[X]$ is the Euler characteristic, and $p_1[X]$ three times the signature. By the Wu formula, $w_2$ is the mod 2 reduction of the coset of $2H^2(X)$ in $H^2(X)$ given by the characteristic vectors, hence is determined by the lattice. In trying to construct an isomorphism of tangent bundles over a given homotopy equivalence, the obstructions one encounters are in $H^2(X;\pi_1 SO(4))=H^2(X;Z/2)$ and in $H^4(X;\pi_3 SO(4))=Z\oplus Z$, and these can be matched up with the three characteristic classes.
Put
$$ R(n)=H^*((S^2)^{\times n}) = \mathbb{Z}[x_1,\dotsc,x_n]/(x_1^2,\dotsc,x_n^2) $$
A key point is that if $u\in R(n)$ with $|u|=2$ then $u^2=0$ iff $u=0$ or $u=m\,x_i$ for some $m\in\mathbb{Z}\setminus\{0\}$ and some $i$; this is easy to check.
Let $\phi\colon R(n)\to R(n)$ be a map of graded rings with $\phi^2=\phi$. Then $\phi(x_i)^2=\phi(x_i^2)=0$ so $\phi(x_i)=0$ or $\phi(x_i)=m\,x_j$ for some $m\neq 0$ and $j$. Using $\phi^2=\phi$ we then get $m(x_j-\phi(x_j))=0$, so $\phi(x_j)=x_j$. Using this, we see that there is a subset $J\subseteq\{1,\dotsc,n\}$ such that $\phi(x_j)=x_j$ for all $j\in J$, and for $i\not\in J$ we have $\phi(x_i)=0$ or $\phi(x_i)=m\,x_j$ for some $j\in J$ and $m\in\mathbb{Z}\setminus\{0\}$. It follows that $\text{img}(\phi)$ is the subring generated by $\{x_j:j\in J\}$, and this is isomorphic to $R(m)$, where $m=|J|$. More precisely, there is an evident inclusion $i\colon(S^2)^{\times m}\to (S^2)^{\times n}$ corresponding to $J$, and the resulting map $R(n)\to R(m)$ restricts to give an isomorphism $\text{img}(\phi)\to R(m)$.
Now suppose we have maps $A\xrightarrow{f}(S^2)^{\times n}\xrightarrow{g}A$ with $g\circ f\approx 1$. Then the map $\phi=(f\circ g)^*$ is an idempotent ring endomorphism of $R(n)$, and $g^*$ induces an isomorphism from $H^*(A)$ to $\text{img}(\phi)$. It follows that there is an inclusion $i\colon(S^2)^{\times m}\to(S^2)^{\times n}$ such that the composite $g\circ i$ gives an isomorphism $H^*(A)\to H^*((S^2)^{\times m})$. As $A$ is a homotopy retract of $(S^2)^{\times n}$ it is also simply connected and of finite type, so we conclude that $g\circ i\colon (S^2)^{\times m}\to A$ is an equivalence.
In particular, we see from this that we cannot have $H^2(A)\simeq H^4(A)\simeq\mathbb{Z}$. Here is a more direct argument for this particular point. In $R(n)$, the product map $R(n)^2\otimes R(n)^2\to R(n)^4$ is surjective, but the squaring map $R(n)^2\to R(n)^4$ is zero mod $2$. If $A$ is a homotopy retract of $(S^2)^{\times n}$, it follows that the product map $H^2(A)\otimes H^2(A)\to H^4(A)$ is surjective, but the squaring map $H^2(A)\to H^4(A)$ is zero mod $2$. This is inconsistent with $H^2(A)\simeq H^4(A)\simeq\mathbb{Z}$.
To push the analysis a bit further, I claim that every ring map $\alpha\colon R(n)\to R(m)$ arises from a continuous map $f\colon (S^2)^{\times m}\to (S^2)^{\times n}$. Indeed, for each $i\leq n$ we must have $\alpha(x_i)=0$ or $\alpha(x_i)=\mu(i)x_{\sigma(i)}$ for some $\mu(i)\neq 0$ and $j\leq m$. If $\alpha(x_i)=0$ then we define $f_i\colon(S^2)^{\times m}\to S^2$ to be the constant map to the basepoint. If $\alpha(x_i)=\mu(i)x_{\sigma(i)}$ then we define $f_i$ to be the composite of the projection $\pi_{\sigma(i)}\colon(S^2)^{\times m}\to S^2$ and a map $S^2\to S^2$ of degree $\mu(i)$. The maps $f_i$ can be combined to give a map $f\colon(S^2)^{\times m}\to(S^2)^{\times n}$, and this has $f^*=\alpha$ as required.
Next, recall that there is a fibration sequence
$$ S^1 =K(\mathbb{Z},1) \xrightarrow{} S^3 \xrightarrow{\eta} S^2 \xrightarrow{} BS^1=\mathbb{C}P^\infty = K(\mathbb{Z},2) \to BS^3=\mathbb{H}P^\infty $$
This gives exact sequences
$$ H^1(X) \to [X,S^3] \to [X,S^2] \to H^2(X) \to [X,\mathbb{H}P^\infty]. $$
Here the first two terms are groups but the last three are only pointed sets. The group $[X,S^3]$ acts on $[X,S^2]$ with stabilisers given by the image of $H^1(X)$ in $[X,S^3]$, and the orbit set maps injectively to $H^2(X)$. Taking $X=(S^2)^{\times m}$ and using $[X,(S^2)^{\times n}]=[X,S^2]^{\times n}$ and noting that $H^1(X)=0$ we arrive at the following conclusion: the group $G(m,n)=[(S^2)^{\times m},(S^3)^{\times n}]$ acts freely on the set $P(m,n)=[(S^2)^{\times m},(S^2)^{\times n}]$, and two maps from $(S^2)^{\times m}$ to $(S^2)^{\times n}$ have the same effect in cohomology iff they lie in the same orbit of this action.
Finally, we note that $S^3=\Omega\mathbb{H}P^\infty$ so $[X,S^3]\simeq[\Sigma X,\mathbb{H}P^\infty]$. Using the standard splitting $\Sigma(X\times Y)\simeq\Sigma X\vee\Sigma Y\vee\Sigma(X\wedge Y)$, we get
$$ [X\times Y,S^3] \simeq [X,S^3] \times [Y,S^3]\times [X\wedge Y,S^3]. $$
Using this repeatedly, we obtain a bijection
$$ G(m,n) \simeq \prod_{i=1}^n [(S^2)^{\times m},S^3] \simeq
\prod_{J\subseteq\{1,\dotsc,m\}}\prod_{i=1}^n[S^{2|J|},S^3].
$$
However, this is not obviously an isomorphism of groups.
Best Answer
The following paper may be helpful:
Wall, C. T. C. Classification of $(nā1)$-connected $2n$-manifolds. Ann. of Math. (2) 75 1962 163ā189.
Let me quote one result from this paper:
This tells you that there is no smoothable fake $S^n\times S^n$ for $n\equiv 6\mod8$