Solution:

The equation requires $x + 2 \gt 0$ and $kx \gt 0,$ and under those restrictions it is equivalent to $kx = (x + 2)^2,$ that is, $x^2 + (4 - k)x + 4 = 0.$

For $k \lt 0$ the restrictions force $-2 \lt x \lt 0.$ On this interval $kx$ decreases from $-2k \gt 0$ to $0$ while $(x + 2)^2$ increases from $0$ to $4,$ so the graphs cross exactly once. Hence every one of the $500$ negative values of $k$ works, while $k = 0$ makes $\log(kx)$ undefined.

For $k \gt 0$ the restrictions force $x \gt 0.$ The quadratic has root product $4,$ so any real roots have the same sign, and the discriminant $(4 - k)^2 - 16 = k(k - 8)$ is negative for $0 \lt k \lt 8.$ When $k \gt 8$ there are two distinct positive roots (root sum $k - 4 \gt 0$), giving two solutions; only $k = 8$ gives exactly one solution, the double root $x = 2.$ In total $500 + 1 = 501$ values of $k$ work.