 Solution

If $p\equiv 1(\mathrm{mod}4)$, choose $i\in {\mathbb{F}}_p$ with $i^2=-1$. The cubic has the three distinct roots $0,im,-im$, so all four two-torsion points are rational over ${\mathbb{F}}_p$. They form a subgroup of order four, hence $4\mid \#E({\mathbb{F}}_p)$ by Lagrange's theorem.

If $p\equiv 3(\mathrm{mod}4)$, let $\chi$ be the Legendre symbol with $\chi (0)=0$. Since $f(x)=x(x^2+m^2)$ is odd and $\chi (-1)=-1$, the terms for $x$ and $-x$ cancel in ${\sum }_x\chi (f(x))$. Counting the possible second coordinates givesThus in both casesThe assumptions $p\ne 2$ and $m\ne 0$ ensure nonsingularity and keep the displayed two-torsion roots distinct when they exist.