SAMC

Looking $\mathrm{mod}4$ we get $(-1{)}^x\equiv (\mathrm{mod}4)$, hence $x$ is even. Looking $\mathrm{mod}3$ we get $1\equiv (-1{)}^z(\mathrm{mod}3)$, 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 $(-1{)}^{a-1}\equiv 1(\mathrm{mod}3)$, 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 $$2^{\alpha }-1=3^t,2^{\alpha }+1=3^s,t+s=m,t<s.$$ 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 $y=2$.