Second Stiefel-Whitney class is a square

At least there are quite a lot of such manifolds: up to multiplying by powers of 2, any oriented bordism class contains such a manifold.

Proof: Let $f: X \to BSO$ be the universal map such that $w_2$ pulled back to $X$ is a square of a class $x$. I.e. $X$ is the fiber of a map $w_2 - x^2: BSO \times K(\mathbb{F}_2,1) \to K(\mathbb{F}_2,2)$. Then there is a Thom spectrum $Mf$ and a map of bordism groups $\pi_d(Mf) \to \pi_d(MSO)$ which becomes an isomorphism after inverting 2. If $d \geq 4$ and $M$ is some oriented manifold, then there exists an $n \gg 0$ such that $2^n [M] \in \pi_d(MSO)$ lifts to $\pi_d(Mf)$, i.e. there is an oriented bordism from the disjoint union of $2^n$ copies of $M$ to a manifold $N$ such that $\tau N: N \to BSO$ admits a lift to a map $N \to X$. If $d \geq 5$ (and maybe 4?), there is no obstruction to doing surgery on $N$ in order to make the map $N \to X$ 2-connected, and hence $H^2(X;\mathbb{F}_2) \to H^2(N;\mathbb{F}_2)$ is injective so $w_2$ stays a non-zero square in $N$. $\square$

If you like, we can partially control the homotopy type of the manifold. For example, if $d \geq 4$ and we write $n = \lfloor d/2\rfloor$ we could use surgery to make the map $N \to X$ $n$-connected. More generally, given an $n$-dimensional finite $CW$-complex $X'$ there is only the "obvious" homotopical obstruction to $X'$ being an $n$-skeleton for a manifold with the requested property: if there exists a map $f: X' \to BSO$ with $f^*(w_2) \in H^2(X';\mathbb{F}_2)$ a non-zero square, then there exists an $n$-connected map from $X'$ to a manifold with the requested property.

First note that the collection of orientable manifolds with $0 \neq w_2(M) = x^2$ for some $x \in H^1(M; \mathbb{Z}_2)$ is closed under products. Moreover, given two such manifolds of the same dimension, their connected sum is another; this is because the Stiefel-Whitney classes of a connected sum are the 'sum' of the Stiefel-Whitney classes of the summands (see here) and $x^2 + y^2 = (x + y)^2$. In addition, given such a manifold, taking a product or connected sum with a spin manifold also provides another such manifold

As $\mathbb{RP}^5$ is an example, the constructions mentioned above provide infinitely many examples in each dimension greater than or equal to five. In dimension four, we would also obtain infinitely many examples if we could find a four-dimensional example, but that is yet to be demonstrated. So the interesting question now becomes:

Does there exist an orientable four-manifold $M$ with $0 \neq w_2(M) = x^2$ for some $x \in H^1(M; \mathbb{Z}_2)$?

Added later: As is shown in this note, if $m \equiv n \bmod 4$, then $w_2(\operatorname{Gr}(m, m + n)) = w_1(\gamma)^2$ which is non-zero except when $m = n = 1$; here $\gamma$ denotes the tautological bundle. As is also shown in the previous link (alternatively, here), the grassmannian $\operatorname{Gr}(m, m + n)$ is orientable if and only if $m + n$ is even. The condition $m \equiv n \bmod 4$ implies $m + n$ is even, so the manifolds $\operatorname{Gr}(m, m + n)$ with $m \equiv n \bmod 4$, except $\operatorname{Gr}(1, 2) = \mathbb{RP}^1 = S^1$, are also examples. In particular, setting $m = 1$, we recover the examples $\mathbb{RP}^n$ with $n \equiv 1 \bmod 4$, $n > 1$.

This observation also provides an explicit four-dimensional example, namely $\operatorname{Gr}(2, 4)$.