I know that in fields of cardinality $p$, $a$ is a quadratic residue if and only if $a^{\frac{p-1}{2}}=1$ (Euler's criterion). Therefore $a^{\frac{p+1}{2}}=a$ and if also $p=3\!\!\!\mod\! 4$ we can exhibit the square roots: $(a^{\frac{p+1}{4}})^2=a$.
Can we do something similar in fields of prime power cardinality (i.e., find square roots of elements with possibly a constraint on the cardinality)? I do not see how trying to do the same thing with the Legendre symbol helps.
This question was marked as a possible duplicate of Quadratic residues in finite field, however, I'm not asking whether an element is a quadratic residue, but assuming it is a quadratic residue, obtain its square roots.
Yes, we can. That criterion and the formula for the square root work because the multiplicative group of integers mod $p$ is cyclic (of order $p-1$). Well, the multiplicative group of the field of cardinality $p^k$ is cyclic as well (of order $p^k-1$).
Thus, for $p>2$, $a$ is a square in the field of order $p^k$ if and only if $a^{(p^k-1)/2}=1$ or equivalently $a^{(p^k+1)/2}=a$. For $p\equiv 3\mod 4$ and $k$ odd, we also have explicit square roots $\left(a^{(p^k+1)/4}\right)^2=a$.
What about $p=2$? In those fields, everything is a square, and the map $x\to x^2$ is a field automorphism.