Find all solutions to the diophantine equation $7^x=3^y+4$ in positive integers. 
Find all solutions to the diophantine  equation $7^x=3^y+4$ in positive integers.
I couldn't have much progress.

Clearly $(x,y)=(1,1)$ is a solution. And there's no solution for $y=2$.
Assume $y \ge 3$ and $x \ge 1$.
By $\mod 9$, we get $7^x \equiv 4\mod 9 \implies x \equiv 2 \mod 3 $.
By $\mod 7$,we get $y \equiv 1 \mod 6$.
I also tried $\mod 2$ but it didn't work.
Please post hints ( not a solution). Thanks in advance!
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A: It's $3(3^a-1)=7(7^b-1)$ with $a=x-1$ and $b=y-1$.
Therefore $7\mid3^a-1$, so $a$ is a multiple of (what?).

Therefore, $3^a-1$ is a multiple of $13$.


Therefore, $7^b-1$ is a multiple of $13$.


Therefore, $b$ is a multiple of (what?).


Therefore, $7^b-1$ is a multiple of $9$.


Therefore, $3(3^a-1)$ is a multiple of $9$.


Therefore, $a$ is (what?).


Therefore, $x$ is (what?).

