, 2019

Clearly, $n = 1$ satisfies the required condition. We now proceed to rule out all integers $n > 1$.

If $n$ is odd, $n \geq 3$, then $n^{n+1} + n - 1$ falls strictly between the squares of two consecutive integers,
$$\left( n^{\left( \frac{n+1}{2} \right)} \right)^2 < n^{n+1} + n - 1 < \left( n^{\left( \frac{n+1}{2} \right)} + 1 \right)^2,$$
so it is not the square, and hence all the less the sixth power of an integer.

Similarly, if $n \equiv 2 \pmod{3}$, then $n^{n+1} + n - 1$ falls strictly between the cubes of two consecutive integers,
$$\left(n^{\frac{(n+1)}{3}}\right)^3 < n^{n+1} + n - 1 < \left(n^{\frac{(n+1)}{3}} + 1\right)^3,$$
so it is not the cube, and hence all the less the sixth power of an integer.

If $n \equiv 0 \pmod{3}$, then $n^{n+1} + n - 1 \equiv -1 \pmod{3}$, so it is not the square, and hence all the less the sixth power of an integer.

Finally, to rule out the only case left, $n \equiv 4 \pmod{6}$, notice that
$$n^{n+1} + n - 1 \equiv (-1)^{n+1} - 1 - 1 \equiv -3 \pmod{n+1}.$$
Since $n+1 \equiv 5 \pmod{6}$, it has a prime divisor $p \equiv 2 \pmod{3}$, $p > 3$, so $-3$ is not a quadratic residue modulo $p$. Consequently, $n^{n+1} + n - 1$ is not a quadratic residue modulo $p$, and hence all the less the square of an integer, let alone the sixth power of one such.