Show that $x^n + x + 3$ is irreducible for all $n \geq 2.$

If we had a decomposition in $\mathbb Z[x]$ $$x^n + x + 3=p(x)\cdot q(x)=(x^a+\dots+c_a)\cdot(x^b+\dots+d_b)$$ we would deduce $c_a\cdot d_b=3$ so that we would have, maybe after renaming the factors, $c_a=\pm 1$ (and of course $d_b=\pm 3$).
This means that for the (complex!) zeros $z_i$ of $p(x)$ we would have $\prod |z_i|=1$, so that at least one of those zeros, say $z_1$, must satisfy $|z_1|\leq 1$.
But then, since $z_1$ is also a zero of our original polynomial: $z_1^n+z_1+3=0$.
This is clearly impossible since $|z_1^n+z_1|\leq |z_1|^n+|z_1|\leq 2$.

This contradiction shows that $x^n + x + 3$ is actually irreducible.

Edit
I now realize that the same method proves that $x^n + x + p$ and $x^n + x - p$ are irreducible for any prime integer $p\geq 3$.

This can be also shown by referring to irreducibility criterion as given by Osada, see for example Theorem 2.2.7 in Prasolov's book Polynomials:

Let $f(x)=x^n+a_1x^{n-1}+\dots+a_{n-1}x\pm p$ be a polynomial with integer coefficients, where $p$ is a prime. If $p>1+|a_1|+\dots+|a_{n-1}|$, then $f$ is irreducible.

Since $3>1+1$, the irreducibility follows.