Let $a := x^2 - 1$ and $b := y^2 - 1$. Then the given equation becomes
$$a^2 b^2 + 16(a+1)(b+1) = z^2.$$
When $a, b > 13$, it is not hard to show that
$$(ab + 8)^2 < z^2 < (ab + 9)^2,$$
and hence there is no solution to the equation. In the case of at least one of $a, b$ is not greater than $13$, or equivalently at least one of $x, y$ is not greater than $3$, one can easily check all the possibilities and conclude that the solutions are $(x, y, z) = (1, n, 4n)$ or $(x, y, z) = (n, 1, 4n)$, where $n \in \mathbb{Z}_{\ge 1}$.