Recall that on a closed $n$-manifold $M$, there is a unique class $\nu_k$ such that $\operatorname{Sq}^k(x) = \nu_kx$ for all $x \in H^{n-k}(M; \mathbb{Z}_2)$; this is called the *$k^{\text{th}}$ Wu class*. The Wu classes can always be written in terms of Stiefel-Whitney classes due to Wu's theorem which states that $w = \operatorname{Sq}(\nu)$.

The first Wu class $\nu_1$ is $w_1$, so $\operatorname{Sq}^1(x) = w_1x$ for all $x \in H^4(M; \mathbb{Z}_2)$ where $M$ is a closed five-manifold. Now, by the Wu formula, we have $\operatorname{Sq}^1(w_3) = w_1w_3$. So $$w_1^2w_3 = \operatorname{Sq}^1(w_1w_3) = \operatorname{Sq}^1(\operatorname{Sq}^1(w_3)) = 0$$ as $\operatorname{Sq}^1\circ\operatorname{Sq}^1 = 0$ by the Ádem relations.

In fact, $w_2w_3$ is the only Stiefel-Whitney number which can be non-zero and therefore defines an isomorphism $w_2w_3 : \Omega_5^O \to \mathbb{Z}_2$. To see this, note that the Stiefel-Whitney numbers of a closed five-manifold are $$w_1^5, w_1^3w_2, w_1^2w_3, w_1w_4, w_1w_2^2, w_2w_3, w_5.$$

It has already been established that $w_1^2w_3 = 0$ and $w_2w_3$ can be non-zero. Note that $w_5 = 0$ as it is the mod $2$ reduction of the Euler characteristic.

One of the properties of Steenrod squares is that $\operatorname{Sq}^k(x) = 0$ if $k > \deg x$. In particular, if we consider $\operatorname{Sq}^k : H^{n-k}(M;\mathbb{Z}_2) \to H^n(M; \mathbb{Z}_2)$, we see that if $k > n - k$ (i.e. $k > \frac{n}{2}$), it must be the zero map and hence by Poincaré duality, we must have $\nu_k = 0$. So on a five-manifold, $\nu_3$, $\nu_4$, and $\nu_5$ must be zero. As is shown here, we have

\begin{align*}
\nu_3 &= w_1w_2\\
\nu_4 &= w_4 + w_1w_3 + w_2^2 + w_1^4\\
\nu_5 &= w_1w_4 + w_1^2w_3 + w_1w_2^2 + w_1^3w_2.
\end{align*}

As $\nu_3 = w_1w_2 = 0$, we have $w_1^3w_2 = 0$ and $w_1w_2^2 = 0$.

Now note that $\nu_5 = w_1w_4 + w_1^2w_3 + w_1w_2^2 + w_1^3w_2 = w_1w_4$ so $w_1w_4 = 0$.

Finally, as $\nu_4$ is zero, so is $w_1\nu_4 = w_1w_4 + w_1^2w_3 + w_1w_2^2 + w_1^5 = w_1^5$.