Let $p$ be any prime factor of $n+1$. Then $p$ is a prime factor of $(n+1)(2n+15)$ and hence of $n(n + 5)$ as well. Since $n(n + 5) = (n + 1)(n + 4) - 4$, we conclude that $p$ divides $4$, and so $p$ can only be $2$. In the same way, we find that the only possible prime divisors of $n+5$ are $2$ and $5$.

Let $n + 1 = 2^a$ and $n + 5 = 2^b 5^c$. Then we have $2^a + 4 = 2^b 5^c$. Note that if $a \ge 5$, the left-hand side is a multiple of $4$ but not a multiple of $8$. Hence we must have $b = 2$ and the equation becomes $2^{a-2} + 1 = 5^c$. As $a \ge 5$, this gives $5^c \equiv 1 \pmod{8}$, forcing $c$ to be even. However, if $c$ is even, then $2^{a-2} = 5^c - 1 \equiv (-1)^c - 1 = 0 \pmod{3}$, which is impossible.

This means that $a$ can only be $0, 1, 2, 3$ or $4$. To find the greatest possible value of $n$, it suffices to show that $a = 4$ is possible. Indeed, if $a = 4$, then $n = 15$, and we have $(n+1)(2n+15) = 16 \times 45 = 2^4 \times 3^2 \times 5$ and $n(n+5) = 15 \times 20 = 2^2 \times 3 \times 5^2$. Therefore, the answer is $15$.