TITLE: Solution in integers to $2^n+n=3^m$
QUESTION [15 upvotes]: 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 $100$.

REPLY [12 votes]: 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.