Saudi Arabian Mathematical Competitions

Looking $\bmod 4$ we get $(-1)^{x} \equiv (\bmod 4)$, hence $x$ is even. Looking $\bmod 3$ we get $1 \equiv (-1)^{z} (\bmod 3)$, hence $z$ is even. Let $x = 2m$, $z = 2n$. We have
$$4^{y} = (5^{n} - 3^{m})(5^{n} + 3^{m})$$
hence $5^{n} + 3^{m} = 2^{a}$, $5^{n} - 3^{m} = 2^{b}$, $a + b = 2y$, $y > 0$. Then
$$2 \cdot 3^{m} = 2^{a} - 2^{b} = 2^{b}(2^{a-b} - 1),$$
so $b = 1$ and obtain
$$3^{m} = 2^{a-1} - 1.$$
Looking mod 3 it follows $(-1)^{a-1} \equiv 1 (\bmod 3)$, hence $a-1$ is even. Let $a-1 = 2\alpha$. We obtain
$$3^{m} = 2^{2\alpha} - 1 = (2^{\alpha} - 1)(2^{\alpha} + 1),$$
and obtain
$$2^{\alpha} - 1 = 3^{t}, \quad 2^{\alpha} + 1 = 3^{s}, \quad t + s = m, \quad t < s.$$
We get
$$3^{s} - 3^{t} = 2$$
or
$$3^{t}(3^{s-t} - 1) = 2,$$
hence $t = 0$, $s = 1$, $\alpha = 1$, $a = 3$.
We get $m = 1$, $5^{n} = 5$, so $n = 1$. Finally, $x = 2$, $z = 2$, and $y = 2$.