Balkan Mathematical Olympiad

We start by observing that $z$ must be even, so $z^2=3^x-5^y\equiv (-1{)}^x-1(\mathrm{mod}4)$ is divisible by $4$, which implies that $x$ is even, say $x=2t$. Then our equation can be rewritten as $(3^t-z)(3^t+z)=5^y$, which means that both $3^t-z=5^k$ and $3^t+z=5^{y-k}$ for some nonnegative integer $k$. Since $5^k+5^{y-k}=2\cdot 3^t$ is not divisible by $5$, it follows that $k=0$ and $$2\cdot 3^t=5^y+1$$ Suppose that $t\ge 2$. Then $5^y+1$ is divisible by $9$, which is only possible if $y\equiv 3(\mathrm{mod}6)$. However, in this case $5^y+1\equiv 5^3+1\equiv 0(\mathrm{mod}7)$, so $5^y+1$ is also divisible by $7$, which is impossible. Therefore we must have $t\le 1$, which yields a (unique) solution $(x,y,z)=(2,1,2)$.