Solution:

Let us call $f(m)=m(m+3)$, and $d(m)$ the number of distinct odd prime factors of $f(m)$. We have that $f(m) f(m+1)=f\left(m^{2}+4 m\right)$; moreover, $f(m)$ and $f(m+1)$ have 2 as their only common prime factor. Indeed, suppose that $p$ is an odd prime dividing both $f(m)$ and $f(m+1)$. Then $p$ also divides $f(m+1)-f(m)=2(m+2)$, and hence $p$ divides $m+2$. But $m+2$ and $m+3$ are coprime, so $p$ divides both $m$ and $m+2$, and we have thus reached a contradiction. It follows that $d\left(m^{2}+4 m\right)=d(m)+d(m+1)$.

Finally, observe that, if the remainders of $d(n)$ and $d(n+1)$ upon division by 3 are distinct and both different from 0, then $d\left(n^{2}+4 n\right)=d(n)+d(n+1)$ is divisible by 3.

We want to show that, for any fixed integer $n>2$, there exists an integer $m \geq n$ such that $d(m)$ is divisible by 3.

We proceed by contradiction: assume that $d(m)$ is not a multiple of 3 for every $m \geq n$. Then, for every $m \geq n$, the remainder of $d(m)$ upon division by 3 must equal the remainder of $d(m+1)$, otherwise $d\left(m^{2}+4 m\right)=d(m)+d(m+1)$ would be a multiple of 3. It follows that $d(m)$ always has the same remainder for every $m \geq n$. This, however, contradicts the fact that $d\left(n^{2}+4 n\right)=d(n)+d(n+1)=2 d(n)$ has remainder 2 if $d(n)$ has remainder 1, and vice versa.