I'm assuming z=x.
x3-4kx2-k2x+4=x(x2-k2)-4(kx2-1).
From this it can be seen that putting k=1 gives us:
x(x2-1)-4(x2-1)=(x-4)(x2-1)=(x-4)(x-1)(x+1), so k=1 is a solution, yielding three roots: 4, 1, -1.
If we use one of these we can reduce the cubic to a quadratic through synthetic division:
1 | 1 -4k -k2 4
1 1 1-4k | 1-4k-k2
1 1-4k 1-4k-k2 | 5-4k-k2=0=(5+k)(1-k), so k=1 (already identified) and k=-5.
The quadratic becomes x2-3k-4=0=(x-4)(x+1), when k=1, which is the solution from above.
It also becomes x2+21x-4=0, but this yields irrational roots. Now try rational zero=2
2 | 1 -4k -k2 4
1 2 4-8k | -2k2-16k
1 2-4k -k2-8k | -2k2-16k+4=-2(k2+8k-2)=0, which has irrational roots.
So it appears that k=1 is the only value of k which produces rational roots.