By $\pmod{3}:$ $\Rightarrow x \equiv 0\pmod{2}$ $x=2k:$ $\Rightarrow 3^y=(5^k-z)(5^k+z)$ $\Rightarrow 5^k-z=3^a, 5^k+z=3^b$ Adding we get that: $5^k=\frac{3^a+3^b}{2}$ $\Rightarrow a=0$ $\Rightarrow z=5^k-1$ $\Rightarrow 3^y+1=2\times 5^k$ $3^1+1=2^2$ $3^2+1=2\times 5$ By Zsigmondy's Theorem: $\Rightarrow \exists$ prime $p\neq 2,5 / p|3^t+1, \forall t>2$ $\Rightarrow y\le 2$ If $y=1$ there is no solution If $y=2 \Rightarrow k=1$ $\Rightarrow x=2$ $\Rightarrow k=4$ $\Rightarrow (x,y,z)=(2,2,4)$ is unique solution $\blacksquare$