Why is it true that if two 4-manifolds are homeomorphic then their squares are diffeomorphic?

It follows by smoothing theory. If $h : X \to Y$ is a homeomorphism between smooth 4-manifolds, one obtains two maps $X \to BO$ which become homotopic in $BTOP$. The difference between them is therefore a map $d: X \to TOP/O$.

Now $TOP/O$ is [No it isn't, see comments] 6-connected, so $d$ is nullhomotopic. Therefore the map $$X \times X \overset{d \times d}\to TOP/O \times TOP/O \overset{\oplus}\to TOP/O$$ is also nullhomotopic, but this is the difference construction applied to the homeomorphism $h \times h : X \times X \to Y \times Y$. It being nullhomotopic means that $h \times h$ is homotopic through homeomorphisms to a diffeomorphism.

EDIT: Out of embarrassment about my mistake, let me try to show something like the opposite: there is a smooth 4-manifold $M$ and a homeomorphism $h : M \to M$ such that $h \times h$ is not isotopic to a diffeomorphism (even though its source and target are the same manifold, so definitely diffeomorphic).

Let $M = S^1 \times \mathbb{RP}^3$. To construct $h$ I will instead construct a topological $s$-cobordism $W$ from $M$ to $M$ and then use the fact that $\pi_1(M) = \mathbb{Z} \oplus \mathbb{Z}/2$ is abelian and hence "good" in the sense of Freedman, so $W$ is homeomorphic to $I \times M$, this homeomorphism yielding $h$.

To construct $W$, use the topological surgery exact sequence, in particular the portion $$\mathcal{S}^{TOP}(M \times I, \partial) \overset{\eta}\to [(M \times I, \partial), G/TOP] \overset{\sigma}\to L_5(\mathbb{Z} \oplus \mathbb{Z}/2)$$ which is an exact sequence of abelian groups. The homotopy type of $G/TOP$ is $K(\mathbb{Z}/2,2) \times K(\mathbb{Z},4)$ up to degree 5, so the middle term is identified with $$H^1(M;\mathbb{Z}/2) \oplus H^3(M;\mathbb{Z})$$ as a group. Furthermore, by chasing through this identification we see that if $[H : W \to M \times I] \in \mathcal{S}^{TOP}(M \times I, \partial)$ is an $s$-cobordism (corresponding to a homeomorphism $h : M \to M$) then the projection to $\eta(H)$ to $H^3(M;\mathbb{Z})$ followed by reduction mod 2 to $H^3(M;\mathbb{Z}/2)$ is precisely the obstruction $\kappa(h) \in H^3(M;\mathbb{Z}/2)$ to $h$ being covered by a map of vector bundles (or of PL-microbundles).

For $M = S^1 \times \mathbb{RP}^3$ we have a class $x \in H^3(M;\mathbb{Z})$ which is i) torsion and ii) reduces to $\bar{x} \neq 0 \in H^3(M;\mathbb{Z}/2)$. Looking at page 171 in Wall's book we find that $L_5(\mathbb{Z} \oplus \mathbb{Z}/2) = \mathbb{Z} \oplus \mathbb{Z}$ (when the orientation character is trivial) so we must have $\sigma(x)=0$, as $\sigma$ is a group homomorphism.

Thus there is a $[H : W \to M \times I] \in \mathcal{S}^{TOP}(M \times I, \partial)$ such that $\eta(H) = 0 \oplus x$, and the associated homeomorphism $h : M \to M$ has $\kappa(h) = \bar{x} \neq 0 \in H^3(M;\mathbb{Z}/2)$.

Finally, $\kappa(h \times h) = \bar{x} \times 1 + 1 \otimes \bar{x} \in H^3(M \times M ;\mathbb{Z}/2)$ is still non-zero, so $h$ is not isotopic to a diffeomorphism.

The original paper of Ivan Smith assumes that the 4-manifolds are simply-connected. So this controls $H^3(X; \mathbb{Z}/2) \cong H_1(X; \mathbb{Z}/2) = 0$.

Possibly this gives an interesting invariant for non-simply-connected 4-manifolds in general?