$\boxed{n=0,1}$.

Obviously, $n$ is non-negative Let $9n+16=k^2$ and $16n=9=m^2$ , where $k,m$ are integers. Thus we have $n=\frac{k^2-16}{9}=\frac{m^2-9}{16}$ which is equivalent to $(4k-3m)(4k+3m)=175$ .....As $175=5\cdot 5\cdot 7$, doing some casework , we end up with $k=m=5$ or $k=4,m=3$, which means $n=0,1$