To solve $2\cdot(5^y)-7^x=1$ How do we find all non-negative integers $x,y$ such that $2\cdot(5^y)-7^x=1$ ? 
Does there exist any solution with $x>2, y>2$ ? 

If $y \geq 3$, then taking $\pmod{125}$, $7^x \equiv -1 \pmod{125}$ so $x \equiv 10 \pmod{20}$. Write $x=10a$ where $a$ is odd. Then $2(5^y)=7^x+1 \equiv 7^{10a}+1 \equiv (-1)^a+1 \equiv 0 \pmod{7^{10}+1}$. However $281 \mid 7^{10}+1$, so we get a contradiction.
Thus there are no solutions with $y \geq 3$. We may check $y=0, 1, 2$ directly.
