Solution:
Call the three equations (1), (2), (3).
(1)/(2) gives y=4z.
(3) + (1) − (2) gives
(y2−z2)x=15z2x=240
so z2x=16.
Therefore
z(x+2z)2=x2z+z2y+4z2x=581z(x−2z)2=x2z+z2y−4z2x=549
so x−2zx+2z=79.
Thus either x=16z or x=4z.
If x=16z, then (1) becomes 1024z3+16z3=1040, so (x,y,z)=(16,4,1).
If x=4z, then (1) becomes 41z3+16z3=1040, so (x,y,z)=(1,16,4).