What are the finite fields for which -1 is not a square?

A square root of $-1$ is an element of order $4$ in the group of units, $\mathbb{F}_q^*$. What do you know about this group, how many elements does it contain, which structure does it have? This will allow you to answer then exactly there is an element $i$ of order $4$. Once you have that, it will be easy to show $i^2 = -1$ as claimed.

Remark: Don't forget about the tricky case of $p=2$. Here, $-1=1 = 1^2$...

Such a field must not contain $\mathbf F_{p^2}$. Hence it is necessarily a field $\mathbf F_{p^r}$ with $r$ odd (and $p\equiv 3\mod 4$), since $\mathbf F_{p^r}\subset\mathbf F_{p^s}$ if and only if $r\mid s$.

Yes, you are right.

Here is how you can prove for $q=p$ prime.

Consider the morphism $x\mapsto x^2\in \mathbb F_p$ for $p\ne 2$, considering its kernel, you can show that there is

$$\frac{p-1}2$$

squares in $\mathbb F_p^*$.

Then let's define

$$X:=\{x\in \mathbb F_p^*,\ x^{(p-1)/2}=1\}.$$

You can show that all squares are in $X$, and since $\mathbb F_p$ is a field, $\vert X\vert \leqslant \frac {p-1}2$.

So $X$ contain all the non-null squares.

And you have:

$$-1\in X\iff \frac{p-1}2\equiv 0\pmod 4\iff p\equiv 1\pmod 4.$$