= Solution
Set $X=Du/\sqrt{1+|Du|^2}$. Since $u\in C^2(\mathbb R^n)$, the field is $C^1$, and $|X|\leq1$ everywhere, even when the <gradient> of $u$ is unbounded. The <divergence theorem> on $B_R(0)$ gives
$$
\kappa|B_R|=\int_{\partial B_R}X\cdot\nu\,dS,\qquad |\kappa|\leq\frac{|\partial B_R|}{|B_R|}=\frac nR.
$$
Letting $R\to\infty$ proves $\kappa=0$. This is the fact that <an entire bounded vector field cannot have nonzero constant divergence>; it holds in every dimension.
The equation is now exactly the <minimal surface equation for a graph>. Apply the supplied <Bernstein theorem for minimal graphs> in dimensions $1\leq n\leq7$ to conclude
$$
\boxed{\kappa=0,\qquad u(x)=a\cdot x+b\quad(a\in\mathbb R^n,\ b\in\mathbb R).}
$$
Thus no nonzero constant right-hand side is possible for an entire graph, and in the stated dimensions every possible graph is an affine plane.
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