Find positive integer $x,y$ such that $7^{x}-3^{y}=4$ Find all positive integers $x,y$ such that $7^{x}-3^{y}=4$.
It is the problem I think it can be solve using theory of congruency. But I can't process somebody please help me .
Thank you 
 A: Let us go down the rabbit hole. Assume that there is a solution with $ x, y > 1 $, and rearrange to find
$$ 7(7^{x-1} - 1) = 3(3^{y-1} - 1) $$
Note that $ 7^{x-1} - 1 $ is divisible by $ 3 $ exactly once (since $ x > 1 $): the contradiction will arise from this.
Reducing modulo $ 7 $ we find that $ 3^{y-1} \equiv 1 $, and since the order of $ 3 $ modulo $ 7 $ is $ 6 $, we find that $ y-1 $ is divisible by $ 6 $, hence $ 3^{y-1} - 1 $ is divisible by $ 3^6 - 1 = 2^3 \times 7 \times 13 $. Now, reducing modulo $ 13 $ we find that $ 7^{x-1} \equiv 1 $, and since the order of $ 7 $ modulo $ 13 $ is $ 12 $, we find that $ x-1 $ is divisible by $ 12 $. As above, this implies that $ 7^{x-1} - 1 $ is divisible by $ 7^{12} - 1 $, which is divisible by $ 9 $. This is the desired contradiction, hence there are no solutions with $ x, y > 1 $.
For an outline of the method I used here, see Will Jagy's answers to this related question.
A: If you take this equation mod 7 and mod 3, it becomes clearer:
$$ 7^x-3^y \equiv 4 \mod 7 \implies -3^y \equiv 4 \mod 7 $$
$$ 7^x-3^y \equiv 4 \mod 3 \implies 7^x \equiv 4 \mod 3 $$
This isolates the variables. You can either use Chinese Remainder Theorem or convert the new modular equations to normal equations to continue (as far as I can see).
