Prove that $2^x = 3 \cdot 9^m+5$ has no positive integer solutions for $m \geq 2$ Prove that $2^x = 3 \cdot 9^m+5$ has no positive integer solutions for $m \geq 2$. 


I noticed that $x \equiv 5 \bmod 6$ and thus $2^x \equiv 4 \bmod 7$, but that doesn't seem to help me since $3 \cdot 9^4 +5 \equiv 4 \bmod 7$. Pretty much any other mod I use doesn't seem to work so I think proof by contradiction or something may work better.
Any solutions must be fairly small - either $m \le 11$ or $x \le 20$.  I assume you have checked all those cases and found no solutions but the two you're already aware of; indeed, there aren't any.
Let's prove this - though with a warning: there's lots of computation and black magic ahead.
EDIT: It turns out we can substantially reduce the amount of computation.  See edit at the bottom.

Suppose $m \ge 12$, and work modulo $3^{12}$.  Since $2$ is a primitive root modulo $3^{12}$, we have a congruence restriction on $x$ modulo $\phi(3^{12}) = 2\cdot3^{11} = 354294.$  In particular, we must have $$x \equiv 199283\pmod{354294}.$$
Now suppose $x \ge 21$, and work modulo $2^{21}$.  $3$ is not quite a primitive root modulo $2^{21}$ - it generates an index-$2$ subgroup of $(\mathbb Z/2^{21}\mathbb Z)^{\times}$, but nonetheless we obtain the restriction $$m \equiv 15627\pmod{524288}.$$
Now let's look for small primes divisible by both 354294 and 524288.  It turns out that $$p = 1+11\cdot\mathrm{lcm}(354294,524288) = 1021636509697$$ is prime, and $11$ is the smallest number for which this is true.

Now consider the sets $$S_2 = \left\{2^{199283+354294\cdot k}\pmod{p}\right\}, \qquad S_3 = \left\{3^{15627+524288\cdot k}\pmod{p}\right\}.$$
It's clear from definitions that $$|S_2| = \mathrm{ord}_p\left(2^{354294}\right) = 720896, \qquad |S_3| = \mathrm{ord}_p\left(3^{524288}\right) = 649539.$$
So we expect approximately $$\frac{720896\cdot649539}{1021636509697} = 0.458\cdots$$ elements in $S_2 \cap (S_3 + 5)$.  If we have no such elements, then there can't be any solutions to $2^x = 3^m + 5$ with $m \ge 12$ and $x \ge 21$.
We get lucky, and there aren't any elements in the intersection!  TA-DA - there are no solutions other than the two you know!

In all seriousness, this shouldn't have worked.  $S_2$ is one-fourth the size we expect; $S_3$ is one-third the size.  And it was still about a fifty-fifty shot if there was anything in their intersection at all.  And $p$ was a factor of two smaller than expected.  We should have had to find much bigger congruences and use those.  But black magic happened, and everything worked out.
One more remark: the reason we chose $p$ divisible by both 354294 and 524288 was that it forced the sizes of $S_2$ and $S_3$ to be fairly small relative to $p$.  But their product should have been roughly $11/2$ the size of $p$ in the worst possible case; as I've said, we got lucky.

EDIT: Here's how to make the numbers involved much smaller.
Work modulo $3^5$; then $x \equiv 23\pmod{162}$.  Work modulo $2^2$; $m$ must be odd.  Then work modulo $163$.  The only possible residue of $2^{23+162k}\pmod{163}$ is $139$, and $134$ is not a possible residue for $3\cdot9^m$ modulo $163$.  Hence we're done.
Suppose that $x$ is a solution, then:
$ 2^{x} = 3.9^{m} + 5 = 3.9^{m} + 3 + 2$, so:
$$2(2^{x-1} - 1) = 3( 9^{m} + 1) \Rightarrow (3.(9^{m} + 1)) | (2^{x-1} - 1)$$
For $m\geq 2$. But $2^{x-1} - 1$ is odd and $3.(9^{m} + 1)$ is pair.