Solution in integers to $2^n+n=3^m$ How to find all positive integers $m,n$ such that $2^n+n=3^m$ ?
We have by inspection $(m,n)=(0,0)$ and $(1,1)$
And there are no more for m and n both less then $100$.
 A: Not a complete answer but some information about possible solutions $n, m$ can be deduced.  Suppose that $n, m > 0$ and write
$$ m = \frac{\log(2^n + n)}{\log(3)}.
$$
Since $\log(x + y) < \log(x) + y/x$ for all real $x, y > 0$ this implies that
$$
\frac{\log(2^n)}{\log(3)} < m < \frac{\log(2^n)}{\log(3)} + \frac{n}{2^n \log(3)}
$$
and therefore that
$$
0 < \frac{m}{n} - \frac{\log(2)}{\log(3)} < \frac{1}{2^n \log(3)}.
$$
So $\tfrac{m}{n}$ is very close to $\tfrac{\log(2)}{\log(3)}$.  So close in fact that it must appear as an upper bound in its continued fraction expansion.  This at least makes it a bit simpler to scan for possible solutions.
A: In this paper Lemma 2.2 gives that if $p,q \in \mathbb{Z}^+$ then
$$
\left|\log_3 2-\frac{p}{q}\right|\ge \frac{1}{1200 q^{14.3}}
$$
limiting how closely $\log_3 2$ can be approximated by rationals.
It then follows that if $m\log 3 = n\log 2 + \epsilon$ with $\epsilon>0$ then $\epsilon\ge \frac{\log 3}{1200 n^{14}}$ and
$$
\begin{align}
3^m & > 2^n (1+\epsilon) \\
& = 2^n + \frac{2^n \log 3}{1200 n^{14}}
\end{align}
$$
The second term is greater than $n$ for $n>112$, and a computer check also excludes $2\le n\le 112$, so there are no other solutions.
