As $b \equiv -a^{2}-3\left(\bmod a^{2}+b+3\right)$, the numerator of the given fraction satisfies
$$a b+3 b+8 \equiv a\left(-a^{2}-3\right)+3\left(-a^{2}-3\right)+8 \equiv -(a+1)^{3} \quad\left(\bmod a^{2}+b+3\right)$$
As $a^{2}+b+3$ is not divisible by $p^{3}$ for any prime $p$, if $a^{2}+b+3$ divides $(a+1)^{3}$ then it does also divide $(a+1)^{2}$. Since
$$0<(a+1)^{2}<2\left(a^{2}+b+3\right)$$
we conclude $(a+1)^{2}=a^{2}+b+3$. This yields $b=2(a-1)$ and $n=2$. The choice $(a, b)=(2,2)$ with $a^{2}+b+3=9$ shows that $n=2$ indeed is a solution.