Fix $\alpha$. The polynomial $P(x):=2^m \alpha^m - (\alpha+x)^m - (\alpha-x)^m$ must be divisible by $3\alpha^2+x^2$. Therefore, $P\left(i\sqrt{3}\alpha\right)=0$. Equivalently, \[2\cos\left(\frac{m\pi}{3}\right) = \left(\frac{1+i\sqrt{3}}{2}\right)^m + \left(\frac{1-i\sqrt{3}}{2}\right)^m = 1\,.\] Therefore, $m \equiv \pm 1 \;(\mathrm{mod}\,6)$. The converse is easily checked. PS: If $m \equiv + 1 \;(\mathrm{mod}\,6)$, we can also see that $P(x)$ is divisible by $\left(3\alpha^2+x^2\right)^2$.