Are there infinitely many elements in this set

For $n \in S$ , let $d$ be the multiplicative order of $3$ mod $2^n$ . Then $d$ divides $n$ by the condition, and $d|2^{n-1}$ by Euler's Theorem.

But $2^n \le 3^d-1$ implies that $d > n \log_3(2) > n/2$ , and since $d|n$ we must have $d = n$ . So $n | 2^{n-1}$ , which implies that $n$ is a power of $2$ . Note that $1,2,4 \in S$ and $8 \notin S$ . We need only show that no higher powers of $2$ are in $S$ , whence the answer will be $\fbox{3}$ .

But this is easy by induction: suppose $n \ge 16$ and write $3^n-1 = (3^{n/2}-1)(3^{n/2}+1)$ . The multiplicity of $2$ in the left factor is $< n/2$ by the inductive hypothesis (base case $n = 16$ corresponds to $n/2 = 8$ which is not in $S$ by inspection), and the multiplicity of $2$ in the right factor is precisely $1$ because it's $2$ mod $8$ . So the multiplicity of $2$ in $3^n-1$ is $< n/2+1 < n$ , and we're done.

The elements in this set are (1,2,4).Therefore the number of elements is 3.