One can easily show that if $n$ is $2$ or $3$, and $a\in\mathbb{Z}$ such that $a$ is not an $n$th power, then $x^n-a$ is irreducible over $\mathbb{Z}$ since it has no roots. However, I believe more generally that:
If $n$ is any positive integer and $a\in\mathbb{Z}$ such that $a$ is not a $p$th power for any prime factor $p$ of $n$, then $x^n-a$ is irreducible over $\mathbb{Z}$.
However, it does not suffice to just check that $x^n-a$ has no roots when $n\geq 3$. So how can one prove this? Is it even true? What if we replace $\mathbb{Z}$ with an arbitrary UFD?
Edit: As per Thomas Andrews's example, assume $\pm a$ is not a $p$th power, not just $a$. In the UFD case, assume no associate of $a$ is a $p$th power.
Let $p(x)\in\mathbb Z[x]$ be an irreducible polynomial which divides $x^n-a.$ By necessity, $p(x)$ is a monic polynomial.
Now, let $d=\deg p(x),$ and assume $d<n.$
Let $\alpha$ be a root of $p(x).$
Show that then $$p(x)=\prod_{\zeta\in J}(x-\zeta\alpha)$$ where $J$ is some subset of the group $G$ of $n$th roots of $1.$
We can show that $J$ is actually the subgroup of size $d,$ and then that we get all the roots by multiplying:
$$x^n-a=\prod_{i=1}^{n/d}\zeta^{-i}p(\zeta^i x)=\zeta^{K}\prod_{i}p(\zeta x)$$
where $\zeta$ is a primitive $n$th root of $1,$ and $K$ is some integer.
But then, substituting $x=0,$ $$-a=\zeta^Kp(0)^{n/d}.$$ That means $\zeta^K$ must be real, hence $\pm1,$ and $\pm a$ is a perfect $n/d$th power.
In particular, if $p\mid n/d,$ then $\pm a=$ is a perfect $p$th power.