Let \(p\) be a prime and \(k,\ell\geq 1\). The field \(\FF_{p^k}\) contains a subfield with \(p^\ell\) elements if and only if \(\ell\mid k\). In that case the subfield is unique, equal to \(\theset{x\in\FF_{p^k}\suchthat x^{p^\ell}=x}\).
If \(F\subseteq\FF_{p^k}\) is a subfield with \(p^\ell\) elements, then \(\FF_{p^k}\) is an \(F\)-vector space of some dimension \(d\), so \(p^k=(p^\ell)^d\) and \(\ell\mid k\).
Conversely, suppose \(\ell\mid k\). Every element of \(\FF_{p^k}\) is a root of \(x^{p^k}-x\), which has degree \(p^k\), so \(x^{p^k}-x=\prod_{a\in\FF_{p^k}}(x-a)\). Since \(\ell\mid k\), \(p^\ell-1\) divides \(p^k-1\), so \(x^{p^\ell-1}-1\) divides \(x^{p^k-1}-1\) and \(x^{p^\ell}-x\) divides \(x^{p^k}-x\). Hence \(x^{p^\ell}-x\) has \(p^\ell\) distinct roots in \(\FF_{p^k}\). They form the fixed set of the field automorphism \(x\mapsto x^{p^\ell}\), which is a subfield with \(p^\ell\) elements. Every element of a subfield with \(p^\ell\) elements satisfies \(x^{p^\ell}=x\), so such a subfield is contained in, and hence equal to, this set.