Solve for integer $m,n$: $2^m = 3^n + 5$ 
Solve for integer $m$ and $n:$ $$2^m = 3^n + 5$$

My Attempt:
Easy to guess two solutions namely $(3,1)$ and $(5,3)$.
Also easy to see that a solution will exist iff $m > 0$ and $n > 0$.
Rewriting it as $2^m - 2 = 3^n + 3$ we get $2^m = 2 \mod 3 \Rightarrow m = 1 \mod 2$ and $3^n = 1 \mod 2 \Rightarrow n = 1 \mod 2$, hence $m$ and $n$ are both odd.
Beyond this, I could not figure out what approach to use.
Source: Past IMO shortlisted problem.
 A: Rename $m\to x$ and $n\to y$
We see $x\geq 3$, $y\geq 1$.
Modulu 3 implies $x$ is odd. For $x\leq 5$ we get only $(3,1)$, $(5,3)$. 
Say $x\geq 6$, then
$$3^y\equiv -5\;({\rm mod}\; 64)$$ It is not difficult to see
$$3^{11}\equiv -5\;({\rm mod}\; 64)$$
so $3^{y-11}\equiv 1\;({\rm mod}\; 64)$. Let
$r=ord_{64}(3)$, then since $\phi(64)=32$,
we have (Euler)
$$3^{32}\equiv 1\;({\rm mod}\; 64)$$
We know $r\;|\;32$. Since
$$3^{32} -1 = (3^{16}+1)\underbrace{(3^8+1)(3^4+1)(3^2+1)(3+1)(3-1)}_{(3^{16}-1)}$$
we get $r=16$ so $16\;|\;y-11$ and thus $y=16k+11$ for some integer $k$. 
Look at modulo 17 now. By Fermat little theorem:
$$2^x\equiv 3^{16k+11}+5\equiv (3^{16})^k \cdot 3^{11}+5\equiv 1^k\cdot 7+5\equiv 12\;({\rm mod}\; 17)$$
Since $x$ is odd we have 
\begin{eqnarray*}
2^1&\equiv & 2\;({\rm mod}\; 17)\\
2^3&\equiv & 8\;({\rm mod}\; 17)\\
2^5&\equiv & -2\;({\rm mod}\; 17)\\
2^7&\equiv & -8\;({\rm mod}\; 17)\\
2^9&\equiv & 2\;({\rm mod}\; 17)
\end{eqnarray*}
so upper congurence is never fulfiled, so no solution for $x\geq 6$.
