II. (50 points) Given a real number
$m$ satisfies: there is one and only one square whose four vertices all lie on the curve $y=x^{3}+m x$. Try to find the area of this square.
Due to the curve $y=x^{3}+m x$ being symmetric about the origin, the center of the square must be the origin (otherwise, reflecting this square about the origin would yield another square with vertices on the curve).

Let one vertex of the square be $A(a, b)$, then $B(b, -a)$, $C(-a, -b)$, and $D(-b, a)$ are the other three vertices. Let the slopes of $O A$ and $O B$ be $k_{1}$ and $k_{2}$, respectively. Since $O A \perp O B$, we have
$$
k_{1} k_{2}=-1,
$$

and (since $A$ and $B$ are on the curve) we have
$$
\begin{array}{l}
k_{1}=\frac{b}{a}=a^{2}+m, \\
k_{2}=-\frac{a}{b}=b^{2}+m.
\end{array}
$$

From (2) and (3), we get
$$
\begin{array}{l}
k_{1}^{2}+k_{2}^{2}=\left(a^{2}+m\right)^{2}+\left(b^{2}+m\right)^{2} \\
=\left(a^{2}+m\right) a^{2}+\left(b^{2}+m\right) b^{2}+\left(a^{2}+m\right) m \\
\quad+\left(b^{2}+m\right) m \\
= a b-a b+k_{1} m+k_{2} m=m\left(k_{1}+k_{2}\right).
\end{array}
$$

Combining (1), we get
$$
\left(k_{1}+k_{2}\right)^{2}-m\left(k_{1}+k_{2}\right)+2=0.
$$

This indicates that the equation
$$
t^{2}-m t+2=0
$$

has a solution $t=k_{1}+k_{2}$. For any solution $k_{1}+k_{2}$ of (6), combined with (1), we can find $k_{1}$ and $k_{2}$ (both real numbers, because $k_{1} k_{2}=-1$). Then, from (2), we determine $a$ and $b$, which satisfy the first equation in (3). From (5), we have
$$
\begin{array}{l}
0=k_{1}^{2}+k_{2}^{2}-m k_{1}-m k_{2} \\
=\left(a^{2}+m\right)^{2}-m\left(a^{2}+m\right)+k_{2}^{2}-m k_{2} \\
=\left(a^{2}+m\right) a^{2}+k_{2}^{2}-m k_{2} \\
=a b+\frac{a^{2}}{b^{2}}+\frac{m a}{b^{2}}=\frac{a}{b}\left(b^{2}+m+\frac{a}{b}\right).
\end{array}
$$

Thus, the second equation in (3) also holds.
Therefore, the four points $A, B, C, D$ determined by $a$ and $b$ form a square on the curve.

Since there is only one square on the curve, the equation (6) has only two identical solutions.
From $\Delta=m^{2}-4 \times 2=0$, we get $m= \pm 2 \sqrt{2}$.
From (2) and (3), we get
$a^{2}+b^{2}=k_{1}+k_{2}-2 m=\frac{m}{2}-2 m=-\frac{3}{2} m$.
Thus, $m=-2 \sqrt{2}$. Hence, $a^{2}+b^{2}=3 \sqrt{2}$.
The side length of the square
$$
A B=\sqrt{2} \quad O A=\sqrt{2} \sqrt{3 \sqrt{2}}=\sqrt[4]{72}.
$$

Therefore, the area of the square $S=A B^{2}=\sqrt{72}=6 \sqrt{2}$.