The Stacks project

Proof. If $X$ is geometrically irreducible over $k$, then $X\times _ k X$ is irreducible by Lemma 33.8.4.

Conversely, suppose that $X \times _ k X$ is irreducible. Replacing $X$ by the underlying reduced scheme, we can assume that $X$ is reduced without changing the problem. Since the first projection $X \times _ k X \to X$ is surjective, $X$ is also irreducible, hence integral. Suppose that there is an element $\alpha $ in the function field $k(X)$ that is separably algebraic over $k$ but not in $k$. Let $E = k(\alpha )$.

Then there is a nonempty affine open subscheme $U$ of $X$ such that $k \subset E \subset O(U)$, and so $E \otimes _ k E \subset O(U)\otimes _ k O(U) = O(U\times _ k U)$. Thus we have a dominant morphism $U \times _ k U \to \mathop{\mathrm{Spec}}(E\otimes _ k E)$. Since $X\times _ k X$ is irreducible, so is $U \times _ k U$, and hence $\mathop{\mathrm{Spec}}(E \otimes _ k E)$ is connected. If $E$ is strictly bigger than $k$, then $E \otimes _ k E$ is a product of fields including $E$ as one factor. It has dimension $(\dim _ k(E))^2>\dim _ k(E)$ as a $k$-vector space, and so $E\otimes _ k E$ is a product of at least two fields, contradicting that $\mathop{\mathrm{Spec}}(E \otimes _ k E)$ is connected. So in fact $E$ is equal to $k$. By Lemma 33.8.6, $X$ is geometrically irreducible over $k$. $\square$