= Solution
If $p\equiv1\pmod4$, 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 $p\equiv3\pmod4$, 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 gives
$$
\#E(\mathbb F_p)=1+\sum_x(1+\chi(f(x)))=p+1\equiv0\pmod4.
$$
Thus in both cases
$$
\boxed{4\mid\#E(\mathbb F_p).}
$$
The assumptions $p\ne2$ and $m\ne0$ ensure nonsingularity and keep the displayed two-torsion roots distinct when they exist.
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