= Solution
Write $Q=q^m$ and take a <splitting field> $L$ of $X^Q-X$ over $F$. Its <formal derivative> is $-1$, since the characteristic divides $Q$, so it has $Q$ distinct <root>[roots] in $L$. Let $K$ be their set. The <Frobenius endomorphism> identity gives
$$
(u+v)^Q=u^Q+v^Q,\qquad(uv)^Q=u^Qv^Q.
$$
Thus $K$ contains $0,1$ and is closed under addition, negatives and multiplication. If $u\neq0$ belongs to $K$, then $(u^{-1})^Q=(u^Q)^{-1}=u^{-1}$, so it is closed under inverses too. Hence $K$ is a <field> with exactly $Q$ elements. Every $c\in F$ satisfies $c^q=c$ by <Lagrange's theorem for finite groups> applied to $F^\times$, together with the zero case; iterating gives $c^Q=c$. Therefore
$$
\boxed{F\subseteq K,\qquad |K|=q^m.}
$$
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