Here is the main result.
Lemma 58.18.1. In More on Morphisms, Situation 37.67.1, for example if $Z \to Y$ and $Z \to X$ are closed immersions of schemes, there is an equivalence of categories
\[ \textit{FÉt}_{Y \amalg _ Z X} \longrightarrow \textit{FÉt}_ Y \times _{\textit{FÉt}_ Z} \textit{FÉt}_ X \]
Proof. The pushout exists by More on Morphisms, Proposition 37.67.3. The functor is given by sending a scheme $U$ finite étale over the pushout to the base changes $Y' = U \times _{Y \amalg _ Z X} Y$ and $X' = U \times _{Y \amalg _ Z X} X$ and the natural isomorphism $Y' \times _ Y Z \to X' \times _ X Z$ over $Z$. To prove this functor is an equivalence we use More on Morphisms, Lemma 37.67.7 to construct a quasi-inverse functor. The only thing left to prove is to show that given a morphism $U \to Y \amalg _ Z X$ which is separated, quasi-finite and étale such that $X' \to X$ and $Y' \to Y$ are finite, then $U \to Y \amalg _ Z X$ is finite. This can either be deduced from the corresponding algebra fact (More on Algebra, Lemma 15.6.7) or it can be seen because
is surjective and $X'$ and $Y'$ are proper over $Y \amalg _ Z X$ (this uses the description of the pushout in More on Morphisms, Proposition 37.67.3) and then we can apply Morphisms, Lemma 29.41.10 to conclude that $U$ is proper over $Y \amalg _ Z X$. Since a quasi-finite and proper morphism is finite (More on Morphisms, Lemma 37.44.1) we win. $\square$
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