Clearly n cannot be a multiple of 3, so the possible choices are n=3k+1, 3k+2
Case 1: if n=3k+1
3|(3k+1)\cdot(2^{3k+1})+1 \equiv 2^{3k+1}+1 \equiv 0 (mod 3)
\Rightarrow 2^{3k+1} \equiv -1 (mod 3)
\Rightarrow 2^{k+1} \equiv -1 (mod 3)
\Rightarrow (-1)^{k+1} \equiv -1 (mod 3)
\Rightarrow k+1 is odd \Rightarrow k is even
therefore k=2m where m \in \mathbb{Z} \Rightarrow n=3(2m)+1=6m+1
2^{3k+1}+1=2^{6m+1}+1 which is always divisible by 3.
Thus we get n=6m+1, m \in \mathbb{Z}.
The second case is very similar to this case.