This is false as stated (ie without some clarification) for all $d>1$. If $m(x)$ is irreducible over $\mathbb{F}_{p^d}$, then it cannot be factored into a product of two polynomials of positive degree. Thus it cannot have any roots in $F$ unless the polynomial is linear, which contradicts the assumption that $d>1$. This has been fixed in an edit.
On the other hand, if you're working over $\mathbb{F}_p$ and $m(x)$ has coefficients all from $\mathbb{F}_p$ and is irreducible over $\mathbb{F}_p$, then the statement is true. Unfortunately, I cannot think of a reason right now why this is true without appealing to the result you are trying to prove.
A different avenue of proof may be the following, if you're familiar with a theorem about splitting fields:
Theorem: Any two splitting fields of a fixed polynomial over a ground field $F$ are isomorphic.
Proof: Exercise.
If you can show that every field of characteristic $p$ contains a subfield isomorphic to $\mathbb{F}_p$ and find a polynomial over $\mathbb{F}_p$ which has exactly $p^n$ distinct roots and find a clever use for the above theorem, you will have a proof.