Suppose $X$ is diffeomorphic to $S^2\times S^2$ or $\mathbb{CP}^2\#\overline{\mathbb{CP}^2}$. Then $X$ is biholomorphic to a Hirzebruch surface.

Note that $b_1(X) = 0$, so $X$ admits a Kähler metric. It follows from Seiberg-Witten theory that any compact Kähler surface with non-negative Kodaira dimension does not admit a metric of positive scalar curvature, but $X$ does, so $X$ has Kodaira dimension $-\infty$. By the classification of surfaces, $X$ is either rational (birational to $\mathbb{CP}^2$) or a ruled surface over a curve of genus $g \geq 1$. The latter possibility cannot occur as $X$ is simply connected, so $X$ is rational and hence biholomorphic to $\mathbb{CP}^2$ or a Hirzebruch surface; see Proposition VI.3.3 of *Compact Complex Surfaces* (second edition) by Barth, Hulek, Peters, and Van de Ven. As $b_2(X) = 2 \neq 1 = b_2(\mathbb{CP}^2)$, we conclude that $X$ is biholomorphic to a Hirzebruch surface.

If $X$ is only homeomorphic to $S^2\times S^2$ or $\mathbb{CP}^2\#\overline{\mathbb{CP}^2}$, then I believe it is still not known whether $X$ must be biholomorphic to a Hirzebruch surface.

As before, $X$ must admit a Kähler metric, but $X$ need not admit a metric of positive scalar curvature a priori, so we cannot immediately determine the Kodaira dimension of $X$ as we did above. Note, if we could show that $X$ had Kodaira dimension $-\infty$, we could conclude that $X$ is biholomorphic to a Hirzebruch surface by the exact same argument. We can however use classification to show that the Kodaira dimension of $X$ is not zero or one.

We can rule out Kodaira dimension zero because any Kähler surface of Kodaira dimension zero is finitely covered by a blownup torus or blownup $K3$ surface, which is not the case for $X$. The impossibility of Kodaira dimension one follows from the fact that $c_1^2 = 0$ for minimal such surfaces, so $c_1^2(X) \in \{0, -1\}$, and $c_2(X) = \chi(X) = 4$, so $\operatorname{Td}(X) = \frac{1}{12}(c_1^2(X) + c_2(X)) \not\in \mathbb{Z}$, which is impossible. Therefore, Hirzebruch's conjecture reduces to providing a negative answer to the following question:

Is there a general type surface homeomorphic to $S^2\times S^2$ or $\mathbb{CP}^2\#\overline{\mathbb{CP}^2}$?

It is natural to ask whether anything changes if $X$ is a compact complex surface which is only homotopy equivalent to $S^2\times S^2$ or $\mathbb{CP}^2\#\overline{\mathbb{CP}^2}$. As $X$ is simply connected, it follows from Freedman's Theorem that $X$ is actually homeomorphic to $S^2\times S^2$ or $\mathbb{CP}^2\#\overline{\mathbb{CP}^2}$, so we're again in the realm of Hirzebruch's conjecture.