Let $(x, y, z)$ be a solution of the problem. Then, notice that $x, y$ are odd, hence $z$ is also odd, and co-prime with $xy$. The equation can be written successively $x^8 + 2x^4y^4 + y^8 = 4z^4$, or $(x^4 - y^4)^2 = 4z^4 - 4x^4y^4$, or $z^4 - (xy)^4 = \left(\frac{x^4-y^4}{2}\right)^2$.

We prove that the equation $a^4 - b^4 = c^2$ $\left(*)\right)$, has no solutions $(a, b, c)$, where $a, b, c$ are positive integers and $a, b$ are co-prime. Assume the contrary to be true. Consider $(a_0, b_0, c_0) \in \mathbb{N}^3$ the solution of the above equation with $a_0$ minimum.

If $b_0$ is odd, from $b_0^4 + c_0^2 = a_0^4$ we deduce that there exist positive integers $m, n, m > n$, of different parities, such that $a_0 = m^2 + n^2$, $b_0^2 = m^2 - n^2$, and $c_0 = 2mn$. It follows that $m^4 - n^4 = (a_0b_0)^2$, i.e. the triple $(m, n, ab)$ is a solution of equation $(*)$, with $m, n, ab > 0$ and $(m, n) = 1$. This contradicts the minimality of $a_0$.

If $b$ is even, there exist $m, n \in \mathbb{N}$, $m > n$, such that $a^2 = m^2 + n^2$, $b^2 = 2mn$, and $c = m^2 - n^2$. We may assume that $m$ is even, $n$ is odd. From $b^2 = 2mn$ it follows that $2m = p^2$, $n = q^2$, i.e. $m = 2p_1^2$, $n = q^2$, with $q$ odd, $(p_1, q) = 1$. Thus, $a^2 = (2p_1^2)^2 + (q^2)^2$, which means that there exist $r, s \in \mathbb{N}$, $r > s$, such that $a = r^2 + s^2$, $2p_1^2 = 2rs$, $q^2 = r^2 - s^2$. From $(r, s) = 1$ and $rs = p_1^2$ it follows that $r = u^2$, $s = v^2$, and $(u, v) = 1$. This means that $u^4 - v^4 = q^2$, i.e. $(u, v, q)$ is a solution of equation $(*)$ with $u, v, q \in \mathbb{N}$, $(u, v) = 1$, and $u < a_0$, which contradicts the choice of $a_0$.

Getting back to the equation
$$z^4 - (xy)^4 = \left(\frac{x^4 - y^4}{2}\right)^2,$$
from the above we can see that the only solutions it may have are with $x^4 - y^4 = 0$, i.e. with $x = y$. But, as $(x, y) = 1$, it follows that one must have $x = y = 1$, and finally that $x = y = z = 1$ is the only solution.