If $F$ is a finite field whose order is a power of 3, then $\sqrt{-1} \in F$ if and only if $\sqrt[4]{-1} \in F$.
My attempt:
Let $\alpha = \sqrt[4]{-1}$.
The converse direction is straightforward, in other words, if $\alpha \in F$ then $\alpha^2 = \sqrt{-1} \in F$.
For the other direction, if $\alpha^2 \in F$, can I adjoin $F$ with $\alpha$, to have $F(\alpha)$ and since $F(\alpha^2) \subset F(\alpha)$ then
$|F(\alpha):F| = |F(\alpha):F(\alpha^2)||F(\alpha^2):F|$.
I am not sure if this leads to the required result.