Diophantine equation: $7^x=3^y-2$ I've tried using mods but nothing is working on this one: solve in positive integers $x,y$ the diophantine equation $7^x=3^y-2$.

Here is a bit ugly solution to this. First, note that $(x,y) = (1,2)$ is a solution. Now suppose $x \geq 2$. We have,
$$
3^y \equiv 2 \pmod{49}.
$$
Now, using Euler's theorem, we deduce 
$$
3^{\phi(49)}\equiv 1 \pmod{49} \implies 3^{42}\equiv 1 \pmod{49}.
$$
Now, we will observe that $3^{21}\equiv (-1)\pmod{49}$, since $3^{21} \equiv (3^{5})^4 \times 3\equiv -1\pmod{49}$.Now with some little algebra we obtain that $3^{26} \equiv 2 \pmod{49}$. Hence,
$$
y\equiv 26\pmod{42}.
$$
Now, we will turn our attention to modulo $43$. With similar tricks again, we observe that
$$
3^{26}\equiv 15 \pmod{43} \implies 7^y \equiv 13 \pmod{43}.
$$
But the values $7^y$ can take in modulo $43$ are given by:
$$
7,6,-1,-7,-6,1
$$
and repeats itself. Hence, $7^y \equiv 13 \pmod{43}$ is not satisfied.
Thus the equation does not have any solution other than the aforementioned one.
