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On a compact five-manifold, the Stiefel-Whitney number $w_2w_3$ can be nonzero.
An example is the manifold $SU(3)/SO(3)$, and also another example is a $mathbbCP^2$ bundle over a circle where the holonomy is given by the complex conjugation automorphism of $mathbbCP^2$.

My question is whether the Stiefel-Whitney number $w_1^2w_3$ can be nonzero on a
compact five-manifold. I haven't found either a simple proof that it is zero, or a simple example where it is nonzero.

Recall that on a closed $n$-manifold $M$, there is a unique class $nu_k$ such that $operatornameSq^k(x) = nu_kx$ for all $x in H^n-k(M; mathbbZ_2)$; this is called the $k^textth$ Wu class. The Wu classes can always be written in terms of Stiefel-Whitney classes due to Wu's theorem which states that $w = operatornameSq(nu)$.

The first Wu class $nu_1$ is $w_1$, so $operatornameSq^1(x) = w_1x$ for all $x in H^4(M; mathbbZ_2)$ where $M$ is a closed five-manifold. Now, by the Wu formula, we have $operatornameSq^1(w_3) = w_1w_3$. So $$w_1^2w_3 = operatornameSq^1(w_1w_3) = operatornameSq^1(operatornameSq^1(w_3)) = 0$$ as $operatornameSq^1circoperatornameSq^1 = 0$ 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 $w_2w_3 : Omega_5^O to mathbbZ_2$. 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 $operatornameSq^k(x) = 0$ if $k > deg x$. In particular, if we consider $operatornameSq^k : H^n-k(M;mathbbZ_2) to H^n(M; mathbbZ_2)$, we see that if $k > n - k$ (i.e. $k > fracn2$), 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

beginalign*
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.
endalign*

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 $w_1nu_4 = w_1w_4 + w_1^2w_3 + w_1w_2^2 + w_1^5 = w_1^5$.