Well, the numbers are small enough for trial and error. The only other way I can think of would be to factor:

x^3 + y^3 = (x + y)(x^2 + xy + y^2) = 35.

The only integer factors of 35 are (1, 35) or (5, 7). There are no integers x and y that add to 1, so x + y = 5 or 7.

Using x + y = 5 we get that y = 5 - x, so y^3 = 125 - 75x + 15x^2 - x^3 so

x^3 + y^3 = 15x^2 - 75x + 125 = 35

or

15x^2 - 75x + 90 = 0

x^2 - 5x + 6 = 0

(x - 3)(x - 2) = 0

So x = 2 or x = 3

Thus y = 3 or y = 2 respectively.

Using x + y = 7 we get that y = 7 - x, so y^3 = 343 - 147x + 21x^2 - x^3 so

x^3 + y^3 = 21x^2 - 147x + 343 = 35

21x^2 - 147x + 308 = 0

3x^2 - 21x + 44 = 0

or

x = (21 +/- sqrt(-87))/6

Since x is complex, this can't be a solution.

So (x, y) = (2, 3) or (x, y) = (3, 2)

However, this is only going to be practical when there are a small number of possibilities for x + y.

-Dan