Let $f(m) = m(m + 3)$ and $d(m)$ be the number of odd distinct prime factors of $f(m)$. We have that $f(m) f(m + 1) = f(m^2 + 4m)$ ; furthermore we see that the only common prime factor of $f(m)$ and $f(m + 1)$ is $2$. Indeed, suppose that $p$ is an odd prime that divides both $f(m)$ and $f(m + 1)$ : then $p$ divides also $f(m + 1) - f(m) = 2(m + 2)$ , so $p$ divides $m + 2$. But $GCD(m + 2, m + 3) = 1$, so $p$ must divide both $m$ and $m +2$ : contradiction. Hence, $d(m^2 + 4m) = d(m) + d(m + 1)$. We see that if the remainders of the division by 3 of $d(n)$ and $d(n + 1)$ are distinct and both non-zero, $d(n^2 + 4n) = d(n) + d(n + 1)$ is divisible by $3$. We want to prove that, chosen any $n > 2$, there exists an integer $m \ge n$ so that $d(m)$ is divisible by $3$. Suppose by contradiction that $d(m)$ is never a multiple of $3$ for any $m \ge n$. Then, for any $m \ge n$ we see that the remainder of $d(m)$ must be equal to the remainder of $d(m + 1)$, otherwise $d(m^2 + 4m) = d(m) + d(m + 1)$ would be a multiple of $3$. Hence, $d(m)$ has always the same remainder for any $m \ge n$. This, however, contradicts the fact that $d(n^2 + 4n) = d(n) + d(n + 1) = 2d(n)$ has $2$ as remainder if $d(n)$ has $1$ as remainder and vice versa, so our thesis is proven.

TL;DR: just bash numbers of odd prime divisors of $6k,6k+1,..,6k+4$ in mod $3$ All congruences here are in mod $3$. Let $\lambda(m)$ denote the number of distinct odd prime factors of $m$. Suppose that there are only finitely many such $m$. Then, there exists $N$ such that for all $m>N$, $\lambda(m(m+3))\not\equiv 0$. Since $3\mid n(n+1)(n+2)$ for all $n$, (1) $\lambda(n)+\lambda(n+1)+\lambda(n+2)=\lambda\left(9n(n+1)^2(n+2)\right)=\lambda\left((3n(n+2))(3(n+1)^2)\right)\not\equiv 0$ for all $n>N$. Choose distinct primes $p,q>3$ such that $6pq>N$. We note which relations of values are impossible: (2) $\lambda(6pq)+\lambda(6pq+1)=\lambda((18pq)(18pq+3))\not\equiv 0$. Similar to the above, (3) $\lambda(6pq+1)+\lambda(6pq+2)\not\equiv -1$ (as $3\nmid (6pq+1)(6pq+2)$) (4) $\lambda(6pq+3)+\lambda(6pq+4)\not\equiv 0$. (5) $\lambda(6pq)+\lambda(6pq+2)=\lambda((9pq)(9pq+3))\not\equiv 0$. (6) $\lambda(6pq)+\lambda(6pq+3)\not\equiv 1$. (7) $\lambda(6pq+1)+\lambda(6pq+4)\not\equiv 0$. Finally, we have (8) $\lambda(6pq)+\lambda(6pq+1)+\lambda(6pq+3)+\lambda(6pq+4)=\lambda((36p^2q^2+24pq)(36p^2q^2+24pq+3))+1\not\equiv 1$. Now since $\lambda(6pq)=3\equiv 0$, from (2), $\lambda(6pq+1) \equiv 1,2$ If $\lambda(6pq+1)\equiv 1$, (1),(3),(5) implies $\lambda(6pq+2)$ cannot be $equiv 2,1,0$ respectively which is impossible. Hence $\lambda(6pq+1)\equiv 2$, and (1),(5) implies $\lambda(6pq+2)\equiv 2$. From (1) and (6), $\lambda(6pq+3)\not\equiv 2,1$ hence $\lambda(6pq+3)\equiv 0$. Now (4),(7),(8) implies $\lambda(6pq+4)\not\equiv 0,1,2$ respectively which is impossible. This is our desired contradiction.

Verify that \[x_{m^2+4m}=(m^2+4m)(m^2+4m+3)=(m)(m+4)(m+1)(m+3)=x_mx_{m+1}.\]Observe that the pairwise difference between any of the factors $m,m+1,m+3,m+4$ is either $1,2,4$, so they share no odd prime factors. Therefore, $\zeta(m^2+4m)=\zeta(m)+\zeta(m+1)$.

Suppose otherwise that for all $n>N$, $3\nmid\zeta(n)$. Then choosing $m>N$, with the relation in the previous claim, we get that under mod 3 the equation must either read $2\equiv1+1$ or $1\equiv2+2$. In other words, $\zeta(m)=\zeta(m+1)$. By induction we get $\zeta$ is constant on all $n$ with $n>N$. But $\zeta(n)\equiv\zeta(n^2+4n)\equiv2\zeta(n)\not\equiv\zeta(n)$, contradiction.