I would like to describe (in a lower level of abstraction) the set $N = \left \{n \in \mathbb{N} : -1 \in \left \langle 2 \right \rangle \leq (\mathbb{Z}/n\mathbb{Z})^x \right \} $.
Clearly $N \subseteq \mathbb{N}_{odd}$. (This set is actually well defined only for odd $n$'s)
Also i heard that $\mathbb{P} \subseteq N$ but not sure how to prove it. (EDIT this is not true, since $-1 \notin \left \{ 1,2,4 \right \} = \left \langle 2 \right \rangle \leq (\mathbb{Z}/7\mathbb{Z})^x$).
Is there more to it? Here are some questions i am trying to answer, yet any other information regarding the set $N$ is wanted:
1) Is $N$ infinite?
2) Can we find a set $M$ such that $N \subset M \subset \mathbb{N}_{odd} $
3) Can we describe the set $N$ more directly?
let's restrict attention to the case when $n=p$ is prime. Then the multiplicative group $(Z/pZ)^\times$ is cyclic and $-1$ is its only element of order $2$. So $-1 \in \langle 2 \rangle$ if and only if $2$ has even order in $(Z/pZ)^\times$.
Now it is well known that $2$ is a square in $(Z/pZ)^\times$ if and only if $p \equiv \pm 1 \bmod 8$.
So if $p \equiv \pm3 \bmod 8$, then $2$ is not a square, and so it must have even order, and hence $-1 \in \langle 2 \rangle$.
If $p \equiv -1 \bmod 8$, then $2$ is a square, and the $(Z/pZ)^\times$ has twice off order, so $2$ has odd order, and $-1 \not\in \langle 2 \rangle$.
I don't know exactly what happens when $p \equiv 1 \bmod 8$. Calculating this for small $p$ suggests that $-1 \in \langle 2 \rangle$ is usually but not always true. It is false for $p=73$, for example.
By the way, I think this question is as much Number Theory as Group Theory, so I will add Number Theory to the tags.