There is an infinite number of solutions of the form (k, -k, 1).

Algebraic manipulation shows 4 more solutions:

c = 1-a-b

c^2 = 1 + a^2 + b^2 + 2(ab - a - b)

a^3 + b^3 + c^2 = a^3+b^3+ 1 + a^2 + b^2 + 2(ab - a - b) = 1

a^3 + b^3 + a^2 + b^2 + 2(ab - a - b) = 0

(a^3 + a^2 - 2a) + (b^3 + b^2 - 2b) + 2ab = 0

a(a+2)(a-1) + b(b+2)(b-1) + 2ab = 0

By inspection if either a or b is zero, then the other (b or a) can be {0, 1, -2} and the equation is satisfied giving 6 solutions, though 2 of them are both (0,0,1) and are included in the infinite set noted above.  The other four:

(0,1,0), (0,-2,3), (1,0,0), (-2,0,3).

There are 2 others that I didn't find until running the program.  I suspect further algebraic manipulation would find them.  [Edit: it looks like broll already did so]

The complete set:

    (a,b,c)

    (k, -k, 1) for all integers k

----------------------

big = 1000

for a in range(-big,big+1):

    for b in range(-big,big+1):

        if a == -b:

            continue

        c = 1-a-b

        if a**3 + b**3 + c**2 == 1:

            print('({},{},{})'.format(a,b,c))