= Solution
\b[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> has
$$
u^2=v^2=z,\qquad uv=0.
$$
These relations follow by using cycles in the separate summands and the degree-one pinch projections onto them. In $N$, the sphere-factor generators $x,y$ satisfy
$$
x^2=y^2=0,\qquad xy=z.
$$
Consequently every degree-two class of $N$ has square
$$
(ax+by)^2=2abz,
$$
an 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 $\operatorname{diag}(1,1)$ and $\left(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right)$, an odd form and an even form respectively.
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