<s>[INST] Solving equation with two unknown degrees 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? 
 [/INST] 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?
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