 Solution

False. Take $M={\mathbb{CP}}^2\#{\mathbb{CP}}^2$ and $N=S^2\times S^2$, with their standard product and connected-sum orientations. They are closed simply connected four-manifolds. For the connected sum, removing a four-ball leaves the fundamental group unchanged, and the Seifert-van Kampen theorem across the separating three-sphere preserves simple connectivity. The Mayer–Vietoris sequence computes integral cohomology $\mathbb{Z},{\mathbb{Z}}^2,\mathbb{Z}$ in degrees zero, two and four, and zero in other degrees. The Künneth theorem gives the same groups for $S^2\times S^2$. Thus their Betti numbers agree and all their integral cohomology is torsion-free.

Their cohomology rings nevertheless differ. In $M$, let $u,v$ come from the two projective summands and let $z$ generate degree four. The cohomology ring of the connected sum of two complex projective planes hasThese relations follow by using cycles in the separate summands and the degree-one pinch projections onto them. In $N$, the sphere-factor generators $x,y$ satisfyConsequently every degree-two class of $N$ has squarean even multiple of the top generator, whereas $u^2$ in $M$ is a primitive generator. An integral graded-ring isomorphism cannot change this property, since it sends a generator of $H^4$ to its positive or negative. A homotopy equivalence induces a graded-ring isomorphism, so none exists. Equivalently, their intersection forms are $\mathrm{diag}(1,1)$ and $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, an odd form and an even form respectively.