<b>Edit:</b> The answer is wrong (Vladimir's comment below gives a counterexample) - it only shows the following: ``Let $f(x) \in \mathbb{F}[x]$ be separable and have no root in $\mathbb{F}$. Assume $\deg(f) \geq 2$ and $d$ is coprime to the characteristic of $\mathbb{F}$. Then $f(x^d)$ is also separable and has no root in $\mathbb{F}$.'' As Loren points out, this is far from guaranteeing irreducibility. 

It seems Jason's comment gives the only exceptions: assume $d$ is prime to $p$ and $\deg(f) \geq 2$. Consider the factorization of $f$ in the algebraic closure of your field $\mathbb{F}$, say $f = \prod_{i=1}^e (x-a_i)$. Since $f$ is irreducible, none of these $a_i$ are in $\mathbb{F}$ (this is where we use $\deg(f) \geq 2$) and since it is separable, $a_i \neq a_j$ for $i \neq j$. Then 

$f(x^d) = \prod_{i=1}^e (x^d-a_i) = \prod_{i=1}^e \prod_{j=1}^d (x-a_{ij})$

where $a_{ij}$'s are $d$-th roots of $a_i$. None of these $a_{ij}$'s are in $\mathbb{F}$ and none of them are equal. <s>So $f(x^d)$ is irreducible as well.</s>