Here is the Vahlen-Capelli irreducibility criterion for binomial polynomials.

**Theorem**. *For a field $F$, nonzero $a \in F$, and integer $n \geq 2$, the polynomial $x^n − a$ is irreducible in $F[x]$ if and only if
the following two conditions are satisfied:
$(i)$ for each prime $\ell$ dividing $n$, $a$ is not an $\ell$th power in $F$,
and $(ii)$ if $4 \mid n$, then $a$ is not $-4b^4$ for some $b \in F$.*

This is proved in Lang's *Algebra*, Theorem 9.1 of Section 9 of Chapter VI. Lang only proves that under those two conditions, $x^n - a$ is irreducible. The converse is easy (you'll need to know $x^4 + 4$ has a universal factorization: it's $(x^2 + 2x + 2)(x^2-2x+2)$).

**Example**. Let $F$ be an arbitrary number field, $n \geq 2$, and $p$ be a sufficiently large prime number. Then $p$ is unramified in $F$, so $p$ is not a $k$-th power in $F$ when $k \geq 2$: if $p = \gamma^k$ for some $\gamma \in F$ then $\gamma \in \mathcal O_F$ and $(p) = (\gamma)^k$ as ideals, but an unramified prime has a squarefree prime ideal factorization. Thus we get a contradiction since $k \geq 2$. So $p$ is not an $\ell$th power in $F$ for each prime $\ell$ dividing $n$. If $4 \mid n$ and $p = -4b^4$ for some $b \in F$ then $(p) = (2b^2)^2$ is the square of an ideal, which is false for large $p$ since $(p)$ is squarefree. Thus the conditions of the theorem are met, so $x^n - p$ is irreducible.

That argument did not need $p$ to be a large *prime*: all the reasoning works for the polynomial $x^n - a$ where $a$ is a sufficiently large squarefree integer.