If $n>5$ or $n<-9$, then: $$n^2(n^2-1)<n^4-8n+15<n^2(n^2+1)$$Now, it suffices to check for the cases where $-9\le n\le5$. Furthermore, we see that the product of two consecutive integers is always even. Therefore, $n$ must be an odd number. This reduces the number of cases to $7$ possible values of $n:-9,-7,-5,-3,-1,1,3,5$ Checking all of these, the only solutions are $n=3,5$.