How to find integers $x,y$ such that $1+5^x=2\cdot 3^y$ Find this equation integer solution
$$1+5^x=2\cdot 3^y$$
I know $$x=1,y=1$$ is such it and $$x=0,y=0$$
I think this equation have no other solution. But I can't prove it.
This problem is from Shanghai mathematics olympiad question in 2014.

If $x=0$, then $y=0$. If $y=0$, then $x=0$.
Let $x,y\ge 1$. Then $1+(-1)^x\equiv 0\pmod{3}$, so $x$ is odd, so $5^x\equiv 5\pmod{8}$, so $3^y\equiv 3\pmod{8}$, so $y$ is odd. Three cases:


*

*$y=3m$. Then $y\equiv 3\pmod{6}$, so $3^y\equiv -1\pmod{7}$, so $5^x\equiv 4\pmod{7}$, so $x=6t+2$, contradiction (because $x$ is odd).

*$y=3m+1$. Then $y\equiv 1\pmod{6}$, so $3^y\equiv 3\pmod{7}$, so $5^x\equiv 5\pmod{7}$, so $x=6t+1$, so $5^x\equiv 5\pmod{9}$, so $3^y\equiv 3\pmod{9}$, so $y=1$, so $x=1$.

*$y=3m+2$. Then $3^y\equiv 9\pmod{13}$, so $5^x\equiv 4\pmod{13}$, impossible.
