bump this

Let $k\geq 3$. Note that $n^k + n + 1 = t^k$ doesnt have any integer solutions since $n^k < t^k < (n+1)^k$. Now take a prime $p$ dividing $n^k + n + 1$ for which $v_p(n^k + n + 1) = m\not\equiv 0 \pmod{k}$ Take $x$ such that $x > p^m$. Note that $v_p(x!) > m$. We have $x! = y^k - (n^k + n + 1)$. Note that $v_p(y^k - (n^k + n + 1))\leq m$ since $v_p(y^k) \equiv 0 \pmod{k}$ and $v_p(n^k + n + 1) = m \not\equiv 0 \pmod{k}$. But $m < v_p(LHS) = v_p(RHS)\leq m$, so there are no solutions to this equation if $x > p^m$. So there are finitely many solutions if $k\geq 3$. If $k=2$, then considering the discriminant of the equation $n^2 + n + 1 = t^2$, we see that only integer solution pairs to this equation are $ (n,t) = (0,\pm 1) , (-1, \pm 1)$ but since $n>0$, this equation doesnt have any solution. So the same argument works here as well.