Closed orientable 4-manifolds with $H_2(M)\cong \mathbb{Z}$ do not admit free actions of $\mathbb{Z}/2$

Let $\phi\colon M\to M$ be a homeomorphism with $\phi^2=\text{id}$. You can use the Lefschetz fixed-point theorem: If $$\Lambda_\phi=\sum_{k\geq0}(-1)^k \text{Tr}(\phi_*|_{H_k(M;\mathbb{Q})})$$ is non-zero then $\phi$ has a fixed-point. Now we have $H_k(M;\mathbb{Q})=H_k(M;\mathbb{Z})\otimes \mathbb{Q}$ and by Poincaré duality together with universal coefficients $H_k(M;\mathbb{Q})\cong H_{4-k}(M;\mathbb{Q})$. The map $\phi$ induces involutions on all these vector spaces, so it is diagonalizable with eigenvalues $\pm1$. With these information you can calculate $\Lambda_\phi\mod 2$ and show that it is non-zero.

A free action of $\mathbb{Z}/2$ leads to a covering map $M \rightarrow M/(\mathbb{Z}/2)$. For an n-sheeted covering map $X \rightarrow Y$, $\chi (X)=n\chi(Y)$. In this case $n=2$ and $\chi(M)$ is odd, by Poincare duality, which yields a contradiction.