The given condition is equivalent to
$$(x + 3)(x^2 + x + 1)f(x) = (x - 1)(x^2 - x + 1)f(x + 1). \quad (1)$$
In equation (1), set
$$x = -3, \frac{-1 - \sqrt{3}i}{2}, \frac{-1 + \sqrt{3}i}{2}, 1$$
respectively, then
$$f(-2) = f\left(\frac{1 - \sqrt{3}i}{2}\right) = f\left(\frac{1 + \sqrt{3}i}{2}\right) = f(1) = 0.$$
In equation (1), set $x = -2, 0$. Then $f(-1) = f(0) = 0$. Hence $-2, -1, 0, 1$ and $\frac{1 \pm \sqrt{3}i}{2}$ are roots of $f(x) = 0$. Then
$$f(x) = (x + 2)(x + 1)x(x - 1)(x^2 - x + 1)g(x), \quad (2)$$
where $g(x)$ is a polynomial with real coefficients. From (2) we obtain
$$f(x + 1) = (x + 3)(x + 2)(x + 1)x(x^2 + x + 1)g(x + 1). \quad (3)$$
Substituting (2), (3) into (1) gives
$$g(x) = g(x + 1).$$
Let $g(x) = \sum_{k=0}^{n} a_k x^k$. Then $\sum_{k=0}^{n} a_k x_k = \sum_{k=0}^{n} a_k (x+1)^k$. Comparing the coefficients of the $(n-1)$-th degree term on both sides, we know
$$a_{n-1} = n a_n + a_{n-1} \Rightarrow n a_n = 0.$$
Therefore, $g(x)$ is a constant $c$. Hence $f(x) = c(x + 2)(x + 1)x(x - 1)(x^2 - x + 1)$, where the constant $c$ is a nonzero integer.

First we prove: $(n + 2)(n + 1)n(n - 1)$ ($n \ge 8$) has at least four distinct prime divisors.
Otherwise, $(n+2)(n+1)n(n-1)$ has at most three distinct prime divisors $2, 3, p$ ($p \ne 2, 3$). But since the greatest common divisors between $(n-1), n, (n+1), (n+2)$ pairwise are $1, 2, 3$, among the two odd numbers, which are coprime to each other, one is $3^a$ and the other is $p^b$, where $a,b$ are positive integers. Consequently, the two even numbers are $2^{c+1}, 2 \times 3^d$, where $c,d$ are positive integers. Hence $|2^c - 3^d| = 1$. Solving this gives $(c,d) = (2,1), (3,2)$.
Therefore, these two even numbers are 8, 6 or 16, 18. The former does not fit. The latter gives the other two odd numbers as 15, 17 or 17, 19, both of which lead to a contradiction.

Next, suppose there exists some positive integer $n$ ($n \ge 8$) such that every prime divisor of $n^2 - n + 1$ is also a prime divisor of
$$(n+2)(n+1)n(n-1),$$
and $(n+2)(n+1)n(n-1)$ has exactly four prime divisors; otherwise, the conclusion holds.
Clearly, $(n^2 - n + 1, n(n+1)) = 1$. From $n^2 - n + 1 = (n+2)(n-3) + 7$, we know
$$(n^2 - n + 1, n + 1) = 1 \text{ or } 3, \quad (n^2 - n + 1, n + 2) = 1 \text{ or } 7.$$
Hence $n^2 - n + 1 = 3^a 7^b$, where $a,b$ are 0 or positive integers. But $9 \nmid (n^2 - n + 1)$, so $a \in \{0,1\}$, and then $b > 0$. By assumption, the prime divisors of $n+2, n+1, n, n-1$ are $2, 3, 7, p$ ($p \ne 2, 3, 7$), so $7|(n+2)$.

Consider the sets of prime divisors $A, B$ of the two even numbers and the two odd numbers respectively among these. Clearly, $2 \in A$, $|B| \ge 2$, $A \cap B \subseteq \{3\}$. So $|A| = 2$ or $|A| = 3$ with $3 \in A$.
If $A = \{2,3\}$ or $\{2,7\}$, then the two even numbers are $2^{c+1}, 2 \times 3^d$ or $2^{c+1}, 2 \times 7^d$, giving
$$|2^c - 3^d| = 1 \text{ or } |2^c - 7^d| = 1.$$
Hence these two even numbers are 16, 18 or 16, 14. The former gives $7 \nmid (n+2)$; the latter makes $(n+2)(n+1)n(n-1)$ have prime divisors $2, 3, 5, 7$ and $13$ (or $17$), a contradiction.

If $A = \{2,p\}$, then $n+2$ is odd, and $n-1$ is even.
$$\text{From } 3 \notin A \Rightarrow 3 \nmid (n-1) \Rightarrow 3 \nmid (n-2).$$
Hence $n+2 = 7^c, n = 3^d$, and $2^e \in \{n+1, n-1\}$ ($c,d,e$ are 0 or positive integers, $c,d \ge 2, e \ge 3$).
Consequently, $|3^d - 2^e| = 1 \Rightarrow (d,e) = (2,3)$. Then $n=9$. Then $n+2 = 11 \ne 7^c$, a contradiction.

If $A = \{2,3,7\}$, then $B = \{3,p\}$, and $n+2$ is even, $(n+2, n-1) = 3$. Hence $2 \times 3 \times 7|(n+2)$.
Consequently, $n = 2^c, n-1 = 3^d, n+1 = p^e$ ($c,d,e$ are positive integers, $c \ge 3, d \ge 2$).
Then $2^c - 3^d = 1 \Rightarrow (c,d) = (2,1)$, a contradiction.

If $A = \{2,3,p\}$, then $B=\{3,7\}$, and $n+2$ is odd, $(n+2, n-1) = 3$. Hence $3 \times 7|(n+2)$. But $(n, n+2) = 1$, so the odd prime divisors of $n$ are not $3,7$, a contradiction.