First let's determine which residues modulo 7 are congruent to cubes of integers:
03 ≡ 0 mod 7
13 ≡ 1 mod 7
23 ≡ 1 mod 7
33 ≡ 6 mod 7
43 ≡ 1 mod 7
53 ≡ 6 mod 7
63 ≡ 6 mod 7.
Therefore the cubic residues modulo 7 are 0, 1, and 6.
These are the only possible values for x3, y3, and z3 modulo 7.
Letting X = x3, Y = y3, and Z = z3, we are looking for solutions to the congruence
X + Y ≡ Z mod 7
with these values.
There are seven solutions:
0 + 0 ≡ 0 mod 7
0 + 1 ≡ 1 mod 7
0 + 6 ≡ 6 mod 7
1 + 0 ≡ 1 mod 7
1 + 6 ≡ 0 mod 7
6 + 0 ≡ 6 mod 7
6 + 1 ≡ 0 mod 7.
Each solution contains a 0, which corresponds to x, y, or z being divisible by 7.
Of course, Fermat's last theorem implies that if x3 + y3 = z3 then at least one of x, y, z is not just divisible by 0 but is equal to 0.
That's a stronger claim than what we've shown.