I am really struggling with this proof. This doesn't seem like it should be that hard. All I have been trying to do is find a $k<3.2^n$ such that $2^k\equiv 1($mod $ 3\cdot 2^n+1)$, but it turns out there are a lot of numbers between $1$ and $3\cdot 2^n$.
I am just not really sure how to go about this otherwise, but I feel like I must be missing something that makes it more rigorous than just guessing until something works.
I also had a go at writing the congruence as $({2^{2^n}})^3-1\equiv0 $ $($mod $ 3\cdot 2^n+1),$ and I then did difference of cubes and got that either ${2^{2^n}}=1,$ which would mean $2$ is not a primitive root, or $({2^{2^n}})^2+{2^{2^n}}+1\equiv0$ $ ($mod $ 3\cdot 2^n+1),$ but then I couldn't work out why the second one can't be zero.
Hint:
By Euler's criterion, $2^{(p-1)/2}\equiv\left(\dfrac2p\right)\pmod p$, and $\left(\dfrac2p\right)=1$ if $p\equiv1\pmod8$.