Solve $2^x \cdot 3^y = 1 + 5^z$ in positive integers.
I think a useful idea for tackling this question is considering congruences. If we consider the equation modulo 4, we get that the RHS is congruent to 2, so $x=1.$ If $y > 1,$ the LHS is congruent to 0 modulo 9. The residues of $5^z$ modulo 9 for $z=1,2,\cdots, 6$ are $5,7,8,4,2,1$. So 5 is a primitive root modulo 9. Thus in order to have $1+5^z\equiv 0\mod 9,$ we must have $z \equiv 3\mod 6$. But then considering the equation modulo 6 or 3 doesn't seem to yield any contradiction.
Best Answer
If $z=6k+3,$ then $5^z+1$ is divisible by $7,$ so it cannot yield a solution.
That finishes your solution, since it means that $x=y=1$ are the only solutions.