Who can prove that a triangular number cannot be a cube, fourth power or fifth power?

This problem was finished off (for arbitrary powers) in a paper of Gyory in 1997 (Acta Arithmetica) : http://matwbn.icm.edu.pl/ksiazki/aa/aa80/aa8038.pdf There are no unexpected solutions. The proof appeals to Darmon and Merel's result on the equation $x^n+y^n=2z^n$ (though, with some care, it should be possible nowadays to prove it using only linear forms in logarithms).

First, notice $n$ and $n+1$ are coprime. And if the product of coprime numbers is a n-th power then both are also n-th powers. Now divide the problem into the cases where $n$ is odd and even. $$n=2t$$ $$t(2t+1)=a^b$$ Then $t$ and $2t+1$ are b-th powers. Let $t=y^b$, $2t+1=x^b$. Then $$x^b-2y^b=1$$ Applying the same substitutions to the case where $n$ is odd you find $$x^b-2y^b=-1$$ In this answer Keith Conrad proves the only solution is $x=1$, $y=0$, which mean $n=0$.