Solving equation with two unknown degrees

Question

I sticked at this problem: $3^x-7^y=2$ where it should be solved for natural $x$ and $y$. I made the conclusion that $x=12a+2$ and $y=6b+1$ after playing with different mods. I am sure that the only solution is $x=2$ and $y=1$ as wolfram alpha confirmed it. Can you help with proof for this?

Answer 1

Let $x>2$ and $y>1$.

Thus, our equation it's $$3^x-9=7^x-7$$ or $$9\left(3^{x-2}-1\right)=7\left(7^{y-1}-1\right),$$ which says that $3^{x-2}-1$ is divisible by $7$, which gives $x-2$ is divisible by $6,$

which gives $3^{x-2}-1$ is divisible by $3^6-1=8\cdot7\cdot13,$ which gives $7^{y-1}-1$ is divisible by $13$,

which says $y-1$ is divisible by $12$, which says $7^{y-1}-1$ is divisible by $7^{12}-1=2^5\cdot3^2\cdot5^2\cdot13\cdot19\cdot181,$

which says $3^{x-2}-1$ is divisible by $19$, which gives $x-2$ is divisible by $18$, which says

$3^{x-2}-1$ is divisible by $3^{18}-1,$ which is divisible by $37$, which gives $7^{y-1}-1$ is divisible by $37$,

which gives $y-1$ is divisible by $9$, which says that $7^{y-1}-1$ is divisible by $7^9-1$, which is divisible by $27$,

which says $9\left(3^{x-2}-1\right)$ is divisible by $27$, which is a contradiction.

Id est, our equation has no solutions for $x\geq3$ and $y\geq2$.

Can you end it now?

Answer 2

Say $y\geq 2$ then $7^y\equiv _{49} 0$ so we have to solve $$3^x\equiv _{49} 2$$

It is not difficult to see $ 3^5\equiv _{49} -2$ and $3^{21}\equiv _{49} -1$, so $$3^{x-26}\equiv _{49} 1$$

and $\color{red}{\rm ord _{49}(3)=42 }$ so $42\mid x-26$ and thus $x=42z+26$ for some integer $z$. Now we have $$3^{26}\cdot (3^{42})^z\equiv _{43} 7^y+2$$

Since $3^{26}\equiv _{43} 15$ and $3^{42}\equiv _{43} 1$ we have to solve $$7^y\equiv _{43} 13$$

Since $7^y \pmod{43}\in \{1,7,6,-1,-7-6\}$ we see that we have no solution for $y\geq 2$.