The Stacks project

Proof. If $X$ is geometrically irreducible over $k$, then it is clear that $X_{k'}$ is geometrically irreducible over $k'$. For the converse, note that for any field extension $k''/k$ there exists a common field extension $k'''/k'$ and $k'''/k''$. As the morphism $X_{k'''} \to X_{k''}$ is surjective (as a base change of a surjective morphism between spectra of fields) we see that the irreducibility of $X_{k'''}$ implies the irreducibility of $X_{k''}$. Thus if $X_{k'}$ is geometrically irreducible over $k'$ then $X$ is geometrically irreducible over $k$. $\square$