analytic number theory - If $2^x$and $3^x$ are integers, must $x$ be as well?

It sure feels like something that should be related to Schanuel's conjecture. Note that this is equivalent to finding integers $m$ and $q$ such that

$$x = frac{ln m + 2pi i q}{ln 3}$$ is not an integer but $2^x$ is. [$3^x$ simplifies to $m$ by construction].

Continuing this, let us actually compute $2^x$. If we split it into real and imaginary parts, a large messy expression ensues. But it naturally splits into 2 reasonable cases, depending on whether $mgt 0$ or $m lt 0$. Let's deal with the positive case first. We get

$$2^x = m^{log_3 2}left( cos(2qlog_3(2) pi) + i sin(2qlog_3(2) pi)right)$$
For that to be an integer, it has to at least be real, but unless $2log_3(2)q$ is an integer, the $sin$ term will not be $0$. For $q=0$, this is $m^{log_3 2}$. We can rewrite that as $2^{log_3 m}$. But we assumed that $m$ was a power of $3$, so $log_3 m$ (and thus $x$) is an integer.

For the $m lt 0$, we get the slightly more complicated

$$2^x = (-m)^{log_3 2}left( cos(2(2q+1)log_3(2) pi ) + i sin(2(2q+1)log_3(2) pi )right)$$
Since $mlt 0$, the first term is real, so we need $2 (2q+1)log_3(2)$ to be an integer for the $sin$ term to disappear, which cannot happen.

Can someone find a flaw in my reasoning? I somehow expect so, as this did not seem difficult, and I would expect it to be if it's an open problem!

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