$5^z \equiv 1 \pmod 6 \Longrightarrow z=2k$ $(5^k-1)(5^k+1)=2^x\cdot 3^y$ $5^k+1 \equiv 2 \pmod 4 \Longrightarrow$ $\begin{cases} 5^k-1=2^{x-1}\\5^k+1 =2 \cdot 3^y \end{cases}$ If k is even then $3 \mid 5^k-1$. If k is odd then $5^k-1 \equiv 4 \pmod 8$ So,$k=1$ $(x,y,z)=(3,1,2)$