Assume $k>1$. Since $\gcd(n, n^2-1)=1$, we can see that $n, n^2-1$ are both $k$th powers. So we have for some $a, b$ that $a^{2k}-1 = b^k$, but it's obvious that for positive $a, b$, no two consecutive $k$th powers can exist. So $k=1$.

As a matter of fact, the product of two or more consecutive positive integers is never a power. For a proof of this fact, see P. Erdös & J. L. Selfridge. The product of consecutive integers is never a power, Illinois J. Math. 19 (1975), 292-301 (MR51 #12692; Zentralblatt 295.10017).

Yes, that's exactly what I had in mind when I first saw this problem. I'm not sure if it would be accepted during the actual competition, though (though technically the Erdos-Selfridge theorem should be citable?), so I wrote up an elementary solution.

Mihailescu's can help here?