This is not always possible.
Suppose you have such a trivialization and suppose $Y$ is connected. Composing with the projection $Y\times_X Z\to Y$, you get a morphism of $X$-schemes $Z\to Y$. It is necessarily étale because $Z, Y$ are étale (SGA 1, Exposé I, cor. 4.8). So $Z\to Y$ is finite and open (by flatness). Its image is thus a connected component of $Y$, hence equal to $Y$. Therefore $Y$ itself is irreducible.
Now there are examples of reducible finite étale covers of irreducible schemes. Let $Y$ be the union of two copies of $\mathbb A^1$ meeting transversally at $0, 1$. Let $\sigma$ be the involution on $Y$ sending $t$ of the first component to $1-t$ on the second component and vice-versa. The action is free and we have a finite étale cover $Y\to X:=Y/\langle \sigma \rangle$
= $\mathbb A^1$. Obviously $X$ is irreducible but not $Y$.
Edit More intuitively, $\sigma$ permutes the components of $Y$ and also permutes the two intersection points.
Edit 2 Sorry, I didn't write what I had in mind ! The quotient of $Y$ by the involution is the affine line with $0$ identified with $1$ (hence a nodal rational curve). This example appears as the reduction of some étale double cover of an elliptic curve with multiplicative reduction.