I think $n$ should be $n\equiv 2\ \ \text{or}\ \ 4\ (mod6)$ Is it ok??

Try again, I corrected a typo in the problem statement.

$mod \ 3$ : $n$ even then $mod \ 7$ : $n \in 2+6 \cdot \mathbb{N}$

$2^k=-1,1$ mod $3$,so $n$ must be even.$2^k=2,4,1$ mod $7$,but $2^{2^n}=2$ mod $7$ if $n$ is even.but $2^2=4,2^4=2,2^6=1,2^8=4,...$ mod $7$,so $2^n=4$ if $n$ is in this form $2+6s$,where $s$ is natural.so finally $n$ is in the form $2+6s$ to divisible by $21$.