I have the direction you mention, so far:
The non-zero elements of $\mathbf F_{q^n}$ are a cyclic group of order $q^n-1=2m$. So, if $y^2=r$, we have $r^m=y^{2m}=1$.
Let $F$ be a field with $q^n$ elements, where $q$ is an odd prime. Write $q^n=2m +1$ with $m \in \mathbb{N}.$
If $r \in F^{\times},$ show that the equation $y^2= r$ has a solution iff $r^m=1.$