Note that $xy(x^2-y^2)$ is always even for all integers $x,y$. Similarly, $yz(y^2-z^2)$ and $zx(z^2-x^2)$ are also always even. Therefore, their sum, $xy(x^2-y^2)+yz(y^2-z^2)+zx(z^2-x^2)$, must also be even. However, $1$ is not an even number, so there are no integer solutions.