Use unique factorization in $\Bbb{C}[x]$. You know that the zeros of the polynomial in $\Bbb{C}$ are $x_k=\zeta^ka^{1/n},$ $\zeta=e^{2\pi i/n}$, $k=0,1,2,\ldots,n-1$. If $f(x)\in\Bbb{Q}[x]$ were a factor of $x^n-a$, then we have
$$
f(x)=\prod_{k\in S}(x-x_k)
$$
for some subset $S\subset\{0,1,2,\ldots,k-1\}$.
What can you say about the constant term of $f(x)$? Remember that it should be rational, and thus have a rational absolute value.
.A little bit of scouting for nice irreducibility criteria throws up some very nice results:
Here is a lovely lemma by (Prof.) Ram Murty:
Let $f(x) = a_mx^m + ... + a_1x + a_0$ be a polynomial of degree $m$ in $\mathbb Z[x]$. Let $H = \displaystyle\max_{0 \leq i \leq m-1} \left|\frac{a_i}{a_m}\right|$. If $f(n)$ is prime for some $n \geq H+2$, then $f(x)$ is irreducible in $\mathbb Z[x]$.
I'll give the link : http://cms.dm.uba.ar/academico/materias/2docuat2011/teoria_de_numeros/Irreducible.pdf
In our case, $a_m = 1$, and the maximum of all the quantities in question is $2$. Hence, if $f(n)$ is prime for some $n \geq 4$, then we are done.
You can check that for $n=4$, the number $f(4) =919$, which is prime!
Hence, it follows that the polynomial is irreducible.
ASIDE : There is also a "shifted" base (base shifts from $0...n-1$ to $|b| < \frac n2$) version of Cohn's criteria, which will tell you that if $f(10)$ is prime, then the given polynomial is irreducible. This matches that description, since all coefficients are between $-5$ and $5$. Very interestingly, $f(10) = 98779$ is also prime! (Hence, another proof by another wonderful result).
Best Answer
The polynomial is already irreducible over $\Bbb F_3$, where it is $$ f=x^4+x^3+x^2+1. $$ So it is also irreducible over $\Bbb Z$ and $\Bbb Q$.
How is this fourth degree polynomial irreducible over $\Bbb F_3$? Because it doesn't have roots in $\Bbb F_3$ and because it is not a product of two irreducible monic polynomials of degree $2$ over $\Bbb F_3$ as well. This is easy here, because we only have $3$ such candidates: $x^2+1,x^2+x−1$ and $x^2−x−1$.