The only possible values in the binary field are 0 and 1. There are two possible values for x, two for y, and two for z so there are (2)(2)(2)= 8 possible values for x, y, and z:
0, 0, 0
0, 0, 1
0, 1, 0
0, 1, 1
1, 0, 0
1, 0, 1
1, 1, 0
1, 1, 1
0, 0, 1
Since your two equations have 4 unknowns, x, y, z, and w, there are 2(2)(2)(2)= 16 possible sets of values but you don't have to look explicitely at all 16.
Since x occurs in both equations, if x= 0, we have w+ z= 0 and y= 1. So we only have to solve w+ z= 0. There are 4 cases:
w= 0, z= 0. Then w+ z= 0+ 0= 0
w= 1, z= 0. Then w+ z= 1+ 0= 1
w= 0, z= 1. Then w+ z= 0+ 1= 1
w= 1, z= 1. Then w+ z= 1+ 1= 0.
Two solutions are x= 0, y= 1, z= 0, w= 0 and x= 0, y= 1, z= 1, w= 1.
If x= 1, then we have w+ 1+ z= 0 and 1+ y= 0. y= 1 is the same as y= "-1"= 1 and w+ 1+ z= 0 is the same as w+ z= "-1"= 1 ("-1" in this field is 1 because 1+ 1= 0).
w= 0, z= 1 and w= 1, z= 0 both give w+ z= 1 so we also have x=1, y= 1, z= 1, w= 0 and x= 1, y= 1, z= 0, w= 1 .
That gives a total of 4 distinct solutions.