51st Ukrainian National Mathematical Olympiad, 3rd Round

We have $y^k=x(x+1)$. If $x=0$ and $x=-1$ then $y=0$.

Suppose that $x\ne 0,-1$.

$x$ and $x+1$ are coprime, therefore $x=a^k$ and $x+1=b^k$ for some integer nonzero $a,b$, that are of the same sign. Then $$(x+1)-x=1=b^k-a^k=(b-a)(b^{k-1}+b^{k-2}a+\cdots +a^{k-1})$$ Which is impossible, since the absolute value of the second bracket is greater than $1$.