Let $k$ be the order of $3$ mod $2^n$. We will prove that for $n \geq 4$ there is: $k > 2^{n-3}$. In fact, we have: $3^{2^{n-3}} \equiv 2^{n-1} + 1 \mod{2^n}$ We prove this inductively. It works for $n=4$. Suppose that it works also for $n \geq 4$, which means that we can write: $3^{2^{n-3}} = a2^n +2^{n-1} + 1$ for $a$ integer after squaring: $3^{2^{n-2}} = b2^{n+1} + 2^{n} + 1$ where $b$ is some integer and we are done. Therefore if $3^n \equiv 1 \mod{2^n}$ then $n \leq 3$ or $n > 2^{n-3}$. Simple checking yields the answer: $n \in \{1, 2, 4\}$.

It may be a known problem, however the source for this is Romania TST 2005 [ See here ], unless someone else has knowledge of an earlier reference.

Use this Lemma: Let $ a, b, n \in \mathbb{Z}$ such that $ 2^{\alpha}\|\frac{a^{2}-b^{2}}{2}$ and $ 2^{\beta}\|n$. Then $ 2^{\alpha+\beta}\|a^{n}-b^{n}$.

Not sure why no one has posted a solution using Lifting the Exponent as this pretty much kills it.

Binomial-theorem wrote: Not sure why no one has posted a solution using Lifting the Exponent as this pretty much kills it.

More simple:

I apologize to everyone for my solution being similar to that of Binomial-theorem. Case1:$n=1$ $n=1$ satisfies the condition. Case2:$n\ge 2$ Since $3^n-1\equiv 0\mod{4}$,then $n$ is even.By Lifting the Exponent Lemma, $v_2(3^n-1)=v_2(3-1)+v_2(3+1)+v_2(n)-1=v_2(n)+2$. Thus $n\le v_2(n)+2$.Let $n=2^kt(k\in \mathbb N, t$:odd$)$.Then $2^k\le 2^kt\le k+2$.This implies that $k=1,2$. Case2-1:$k=1$ $2t\le 3\rightarrow t=1\rightarrow n=2$ which satisfies the condition. Case2-2:$k=2$ $4t\le 4\rightarrow t=1\rightarrow n=4$ which satisfies the condition. From above cases,the answer is $\boxed{n=1,2, 4}\blacksquare$