For $p$ prime, $P(x)=1+x+...+x^{p-1}$ is irreducible in $\mathbb{Z}[x]$.
This is a classic problem to which there exists a clever solution which applies Eisenstein's criterion to $P(x+1)$.
However I believe I have another solution, but I wish to make sure I haven't made some stupid mistake:
We have $P(x)(x-1)=x^p-1$. For $f$ a polynomial in $\mathbb{Z}[x]$, let $\overline{f}$ denote it's reduction mod $p$, which is a polynomial in $\mathbb{F}_p[x]$.
By Fermat, we have that $\overline{P(x)(x-1)}=x-1$ so $\overline{(P(x)-1)} \overline{(x-1)}=0$. But $\mathbb{F}_p[x]$ is an integral domain so $\overline{P(x)}=1$. Thus if $P=QR$ for nonconstant polynomials $Q,R$ in $\mathbb{Z}[x]$ then $\bar{Q}\bar{R}=1$. Hence $\bar{Q}$ and $\bar{R}$ are constants polynomials. Thus the leading coefficients of $Q$ and $R$ are divisible by $p$, which means the leading coefficient of $P$ is divisible by $p$, a contradiction.
This does not quite work; if it did, the same logic should hold for $x^{p^p}-1$; for instance, say $x^{27}-1$. This should imply $$ 1+x+\dots+x^{26} $$ is irreducible, but it is not, as can be checked by wolfram. The problem is as @Wojowu points out in the comments.