Factoring in $\mathbb{Z}[x]$ vs in $\mathbb{Q}[x]$

Let $\mathbb{Q}[x]$ be the ring of polynomials in the variable $x$ with rational coefficients and $\mathbb{Z}[x]$ the ring of polynomials in the variable $x$ with integer coefficients.

Let $f(x) = 2 + x + \displaystyle \sum_{k = 2}^{2026} (x+1)^k$ .

In which ring(s) of polynomials is $f(x)$ irreducible? (irreducible basically means, you can't factor it into a product of lower degree polynomials. More details below.)

NB: Some terminology which may be helpful:

A set $R$ equipped with the abstract operations $+, \times$ is a ring if the following criteria are satisfied:

a. $R$ is an additive abelian group.

b. $\forall a, b, c \in R, a(b+c) = ab + ac; (a+b)c = ac + bc$

c. $\exists 1 \in R$ such that $\forall r \in R, 1r = r1 = r$

Example. $\mathbb{Z}$ under $+, \times$ is a ring.

Let $R$ be a ring. $r$ is a unit of $R$ if there exists $s \in R$ such that $rs = sr = 1$ .

Example. $3$ is a unit in $\mathbb{Z}/5 \mathbb{Z}$ , also in $\mathbb{Q}$ , but not in $\mathbb{Z}$ .

Let $R$ be a ring. $r \in R$ is a zero divisor if there exists nonzero $s \in R$ such that $rs = 0$ .

Example. In $\mathbb{Z}/6\mathbb{Z}$ , $2$ and $3$ are nonzero zero divisors. In any ring, $0$ is always a zero divisor. In $\mathbb{Z}/5 \mathbb{Z}$ , there are no nonzero zero divisors.

Let $R$ be an integral domain (that is, there are no nonzero zero divisors in $R$ ). We say that a nonzero, non-unit polynomial $f$ in the polynomial ring $R[x]$ over the ring $R$ is irreducible if $\forall g, h \in R[x]$ such that $f = gh$ , either $g$ is a unit in $R[x]$ or $h$ is a unit in $R[x]$ .

Example. Any linear polynomial (degree 1) is irreducible.