$b | (n+2)a^2-(n+1)a-n$ Set $n=b$ then $b|2a^2-a \to b|2a-1$ and as $a<b$ then $b=2a-1$ $n=b-2: b|a+2 \to b=5,a=3$ $(n+2)3^2-(n+1)*3-n=5n+15$ so $(a,b)=(3,5)$ is answer

The expression equals $a^{n+1000}((n+2)a^2 - (n+1)a - n)$ and since $a$ and $b$ are coprime, we need to work only with $b$ dividing $(n+2)a^2 - (n+1)a - n$. In particular, $n=1$ and $n=2$ give $3a^2 - 2a-1$ and $4a^2 - 3a-2$, so (after some manipulations) $b$ divides $15$ and only $b=5$ (with $a=3$) works (easy to check explicitly).