Exponential Diophantine equation $7^y + 2 = 3^x$ 
Find all positive integer solutions to $$7^y + 2 = 3^x.$$


ATTENTION: MY SOLUTION HAS A TERRIBLE MISTAKE WHICH I HAVE OVERLOOKED!
Obviously, $x > y$. Then, we have $3^x = 7^y + 2 \equiv 0 \pmod {3^y}$. Also, $$7^y = (6 + 1)^y = \sum_{k = 0}^{y} {y \choose k} 6^k \equiv \sum_{k = 0}^{y - 1} {y \choose k} 6^k \pmod {3^y}.$$ We claim that the highest power of $3$ that divides ${y \choose k}$ is at most $2$. Indeed, $$\sum_{i = 1}^{\infty} \left [\frac {y} {3^i} \right] - \left (\sum_{i = 1}^{\infty} \left [\frac {y - k} {3^i} \right] + \sum_{i = 1}^{\infty} \left [\frac {k} {3^i} \right] \right) \leqslant 2.$$ Hence, $$7^y \equiv \sum_{k = 0}^{y - 1} {y \choose k} 6^k \leqslant 2 \sum_{k = 0}^{y - 1} 6^k = \frac {2} {5} (6^y - 1).$$ Since $(5, 3^y) = 1$, we have by Euler's Theorem that $$5^{\phi (3^y)} = 5^{3^y - 3^{y - 1}} \equiv 1 \pmod {3^y}.$$ Then, $2 \cdot 3^{y - 1} \equiv \frac {2} {5} (6^y - 1) \pmod {3^y}$ and $$0 \equiv 7^y + 2 \leqslant 2 \cdot 3^{y - 1} + 2 \pmod {3^y}.$$ Take $s > 0$ an integer for which $$3^{y} s \leqslant 2 \cdot 3^{y - 1} + 2.$$ It follows from this that $0 \leqslant 3^{y} (s - 1) \leqslant 2 - 3^{y - 1}$. Hence, $y < 2$. So the solutions are $$(x, y) = (1, 0), (2, 1).$$
"Notes on Olympiad Problems", Nima Bavari, Tehran, 2006.

Okay so I got this solution after modular bashing for an hour. This better be right.
Since the cases $x,y \le 2$ are already investigated above easily, we look at $x,y \ge 3$.
Rewrite this equation to $$7(7^{y-1}-1)=9(3^{x-2}-1)$$
Now, since $7|3^{x-2}-1$, and the order of $3$ modulo $7$ is $6$, we have $6|x-2$.
This gives $13|3^6-1|3^{x-2}-1$, so $13|7^{y-1}-1$. 
Now we have $12|y-1$ since the order of $7$ modulo $13$ is $12$. This gives $19|7^{12}-1|7^{y-1}-1$.
Now we have $19|3^{x-2}-1$, so $18|x-2$, since the order of $3$ modulo $19$ is $18$.
Now $37|3^{18}-1|3^{x-2}-1$. This gives $37|7^{y-1}-1$.
Now we have $9|y-1$. Now we have $27|7^9-1|7^{y-1}-1$, since the order of $7$ modulo $37$ is $9$.
However, $9(3^{x-2}-1) \equiv -9 \pmod{27}$, so it cannot be a multiple of $27$.
We now have a contradiction, so the answer is $(x,y)=(1,0),(2,1)$, as desired. GG!!
