Elementary solution of exponential Diophantine equation $2^x - 3^y = 7$. The title says it all. I would like to have a solution, preferably one which is as elementary as possible, of the exponential Diophantine equation
$$
2^x - 3^y = 7
$$
where $x,y$ are non-negative integers. Note that some small solutions are $(x,y)=(3,0)$ and $(x,y)=(4,2)$. If I really had to solve it at all costs, I would translate this to the problem of finding integral points on a bunch of curves of genus $1$. However, I would like to know if there are any simpler methods out there.
As far as I can see, simple congruence tricks won't work: $2^x = 7$ is soluble $3$-adically and $-3^y = 7$ is soluble $2$-adically, so I can't see how we could get anything by looking $p$-adically for $p=2$ or $p=3$, and I think the fact that the solution set to the original problem is non-empty means that $p$-adic considerations for $p \neq 2,3$ have no chance of working either. (But maybe I'm wrong.) 
 A: Looking at the equation modulo $ 3 $ gives that $ 2^x \equiv 1 \pmod{3} $ unless $ y = 0 $, hence $ x $ is even. On the other hand, modulo $ 7 $ we have $ 2^x \equiv 3^y \pmod{7} $, and since $ 2 \equiv 3^2 \pmod{7} $ and $ 3 $ is a primitive root modulo $ 7 $, this implies that $ 2x - y $ is divisible by $ 6 $, and hence $ y $ is even also. Writing $ x = 2m $ and $ y = 2n $, we find
$$ 2^{2m} - 3^{2n} = (2^m - 3^n)(2^m + 3^n) = 7 $$
Now, we use the primality of $ 7 $, and it is easily seen that the only solution is $ m = 2, n = 1 $. If $ y = 0 $, then obviously $ x = 3 $, so the only solutions are $ (4, 2) $ and $ (3, 0) $.
A: Another slightly simpler try avoiding primitive roots as used by @Starfall:
$$\begin{array} {ccl} 2^x &- 3^y &= 7 \\
 2^x &&\equiv 1 &\pmod 3 &\implies x=2x_1 \\
 4^{x_1} &- 3^y &= 7 \\
  &       - 3^y &= -1 &\pmod 4  &\implies y=2y_1 \\
 4^{x_1} &- 9^{y_1} &= 7 \\
 4^{x_1}& & \equiv 7 &\pmod 9 &\implies x_1=2 + 3x_2 \\
 2^{2(2+3x_2)} &- 3^{2y_1} &= 7 \\\end{array}$$
and then, because the exponents are even, by factoring and using that $7$ is prime:           
$\qquad \qquad \displaystyle\begin{array}{rcc}
 \underset{a=1}{\underbrace{(4 \cdot 8^{x_2} - 3^{y_1})}}&\cdot& {\underset{b=7}{\underbrace{(4 \cdot 8^{x_2} + 3^{y_1})}}} &= 7 &\qquad \qquad&&&\\
\end{array} $         
and finally       
$\displaystyle \qquad \qquad b=7 \implies x_2=0, y_1=1 \\
\qquad \qquad \phantom {b=7}\implies x=4, y=2 \qquad \text{ is the only solution}$               
