When is $3^n + n$ a power of 2? For what $n \in \mathbb{N}$ is $3^n + n$ a power of $2$?
 A: This is not a complete solution, but shows that any nontrival solutions must be mindbogglingly big if it exists at all.
We have the trivial solutions $3^0+0=2^0$ and $3^1+1=2^2$.
Assume $3^n+n=2^m$ with $n\ge 2$. Then clearly $m>n$.
We find $1+n\equiv 0\pmod 2$, hene $n$ is odd. Then $3^n\equiv 3\pmod 8$ and hence $n\equiv 5\pmod 8$, esp. $m\ge 6$.
From $3^8\equiv 1\pmod{32}$ we have $3^n\equiv 19\pmod{32}$, hence $n\equiv 13\pmod {32}$. A pattern emerges.
Lemma: For $k\in\mathbb N$ we have $3^{2^k}\equiv 1\pmod {2^{k+2}}$
Proof: This is true for $k=1$ and from
$$3^{2^{k+1}}-1=(3^{2^k}-1)(3^{2^k}+1) $$
the claim follows by indcution because $3^{2^k}+1$ is even. $_\square$
Propositio: Assume for some $1\le k< a<2^k$ we have that for all nontrivial solutions of $3^n+n=2^m$ we have $n\equiv a\pmod{2^k}$.
Then  for all nontrivial solutions of $3^n+n=2^m$ we have $n\equiv -3^a\pmod{2^{k+2}}$.
Proof: Let $3^n+n=2^m$ be a nontrivial solution.
Then with $n=2^kb+a$ for some $b\in\mathbb N_0$ and by the lemma
 $$ 3^n=3^a\cdot (3^{2^k})^b\equiv 3^a\pmod {2^{k+2}}.$$
As $m>n\ge a>k$ we find $3^a+n\equiv 0\pmod{2^{k+2}}$. $_\square$
Using the proprosition we can start with $(k,a)=(3,5)$ and repeatedly replace this with $(k+2, -3^a\bmod {2^{k+2}})$. The process either ends with a pair $(k,a)$ with $a\le k$ (and then necessarily $a=\in\{k,k-1\}$) or it never ends. In the latter case we conclude that no nontrivial solution exists, in the former case we may have found a solution, and if we give upprematurely, we at least obtain an estimate and modular condition for all nontrivial solutions. The sequence starts 
$$ (3,5), (5,13), (7,45), (9,173), (11,685), (13,685), (15,25261)$$
and after a few more steps one reaches
$$(k,a)=(201,864075976670532385554180581999784042802808809920656868008621)$$
Especially, $m>n>8.64\cdot 10^{59}$. Also we can continue at leat until $k\approx 8.64\cdot 10^{59}$ and expect $a$ to grow accordinglyet.c sothat the sequence never end and presumably no solution exists.
By taking logarithms, we also find that $\frac{m}{n}$ is an extremely good approximation to $\frac{\ln 3}{\ln 2}$, which is also a hint towards non-existence of a solution.
