I put the eq. in the folowing form
Y^2 - x^3 = 432
So i divided by prime numbers 432 = 2^4*3^3, then the result must be multiple of two's and three's. Maybe there is a scientific method, I don't know it. I think it was lucky.
Note that x3 = y2 + 432 is a perfect cube 63(y2 + 432) = 216(y2 + 432) is a perfect cube. But 216(y2 + 432) = (y + 36)3 - (y - 36)3.
Hence (6x)3 + (y - 36)3 = (y + 36)3. (1)
By Fermat's Last Theorem, an + bn = cn has no non-zero integer solutions for a, b and c, when n > 2. Here we need the result only for the case n = 3, which was first proved by Euler, with a gap filled by Legendre.
However, x > 0.
Hence (1) can hold only when y - 36 = 0 or y + 36 = 0; that is, y = ±36, in which case 6x = 72.