You correctly recalled the extension to the law of of quadratic reciprocity. So in a prime field $\Bbb{F}_p=\Bbb{Z}/p\Bbb{Z}$ $2$ is a square if and only if $p\equiv\pm1\pmod8$.
In an extension field $\Bbb{F}_q$, $q=p^n$, $p$ an odd prime, you can think of it as follows. If $p\equiv\pm1\pmod8$, then $2$ is a square already in the prime field $\Bbb{F}_p\subseteq\Bbb{F}_q$. But if $p\equiv\pm3\pmod8$, then the polynomial $x^2-2$ is irreducible over $\Bbb{F}_p$. Therefore $K=\Bbb{F}_p[x]/\langle x^2-2\rangle$ is a field, and the zeros of $x^2-2$ both (they are negatives of each other) exist in $K$. We see that $[K:\Bbb{F}_p]=2$, so $K$ has $p^2$ elements. Because to each prime power $q$ there is a unique field of cardinality $q$ we see that $2$ is a square in the field $\Bbb{F}_{p^2}$ when $p\equiv\pm3\pmod8$.
So in the case $p\equiv\pm3\pmod8$ we see that $2$ is a square in the field $\Bbb{F}_{p^n}$ if and only if $K\subseteq\Bbb{F}_{p^n}$. By one of the other basic results about finite fields we know that $\Bbb{F}_{p^2}\subseteq\Bbb{F}_{p^n}$ if and only if $2\mid n$.
Summary: $2$ is a square in the field $\Bbb{F}_{p^n}$, $p$ odd, if and only if either $p\equiv\pm1\pmod8$ or $n$ is even.