jmc

Since $g(x)$ is divisible by $x-4$, we have $g(4)=0$. We also have $$\begin{array}{rl}g(4)& =4^3-4^2-(m^2+m)(4)+2m^2+4m+2\\ & =50-2m^2,\end{array}$$so $0=50-2m^2$. Thus $m$ can only be $5$ or $-5$. We check both possibilities.

If $m=5$, then $g(x)=x^3-x^2-30x+72=(x-4)(x^2+3x-18)=(x-4)(x+6)(x-3)$, so all zeroes are integers.

If $m=-5$, then $g(x)=x^3-x^2-20x+32=(x-4)(x^2+3x-8)$, but $x^2+3x-8$ does not have integer zeroes.

Therefore, the only solution is $m=\boxed{5}$.