A field in which every element (that is not 1 or 0) is a root of -1 Let $\mathbb{F}$ be a field with $char(\mathbb{F}) \neq 2$ such that for every element $q \in \mathbb{F}$ if $q \neq 0$ and $q \neq 1$ then there is a power n such that $q^n = -1$. (E.g. $\mathbb{F}_3$, $\mathbb{F}_5$, $\mathbb{F}_{9}$, $\mathbb{F}_{17}$, ...).
Obviously $char(\mathbb{F}) > 0$ (since $\mathbb{F}$ cannot have a $\mathbb{Q}$ subfield). 
Furthermore, I guess in finite fields this happens if and only if $|\mathbb{F}| = 2^k+1$ for some k (since in this case $\mathbb{F}^*$ is a cyclic group of order $2^k$ and -1 lies in every non-trivial subgroup)
Of course $\mathbb{F}$ is not algebraically closed (since every element has even multiplicative order and thus $x^{(2n+1)} - 1$ has only 1 root $\forall n$).
But can $\mathbb{F}$ be infinite? 
Also is there an infinite number of (finite) fields with this property?
 A: The number of such fields is likely finite (up to isomorphy) but it is not know unconditionally. 
The complete list can be described as: the fields with cardinality a Fermat prime, i.e., a prime of the form $2^k+1$, or $9$.  (The former is likely finite yet noone knows.)
As you observed correctly the condition for a finite field is that the order of $|F^{\times}|$ is a power of two. 
To see this it suffices to note that $-1$ always has order $2$ in this group, so if there is an element of odd order in this group then no power of it can have order $2$. Thus the order of the group cannot be divisible by any odd prime. Conversely, as you said if the order is a power of two then $-1$, the unique element of order $2$, is contained in every non-trivial subgroup and thus a power of it is $-1$. 
Now, it is unknown if there are infinitely many primes of the form $2^k +1$, but the answer is likely no. 
What about prime powers, there the only solution is $3^2$; this is a consequence of Catalan's conjecture (proved by Mihailescu), but that special case was known before. 
Clearly no field containing a transcendental element can have your property, and every infinite algebraic field will contain as subfield a finite field not in our list. 
