Smallest possible based on Fermat's Last Theorem of n = 3

Correct me if I'm wrong, but I think I have a proof that (n,s) = (0,1) is the only integer solution to the equation. Here's what I did.... $\ 4(3n+1)^3 \equiv0\ mod(4) \Rightarrow\frac{4(3n+1)^3 -1}{3} \equiv1 \ mod(4)$ $\therefore\ S \ is \ odd$ Let S = 2m+1: $\ 3(2m+1)^2 +1 = 4(3n+1)^3$ $\Rightarrow\ 3m^2 +3m+1 = 3m(m+1) +1 = (3n+1)^3$ Now m(m+1) is even as it is the product of two consecutive integers, so by parity (odd +even = odd), $\ (3n+1)^3$ is odd. Hence, n must be even so let n = 2 $\phi$ , then: $\ 3m^2 +3m+1 = (6 \phi +1)^3 = 216 \phi^3 +108 \phi^2 +18 \phi +1$ $\Rightarrow\ m^2 +m = m(m+1) = 6 \phi (12 \phi^2 +6 \phi +1)$ Now by using the discriminant ( $\ b^2 -4ac \geq\ 0$ ) we see that the quadratic factor has no real roots so $\ m = 6 \phi \ and \ m+1 = 12 \phi^2 +6 \phi +1$ (It doesn't work the other way around as $\ 12 \phi^2 +1 \geq\ 1$ ) This implies that $\phi = 0$ so m = 0 or -1. This gives n= 0 and S = 1 (S = -1 is not valid as S is a natural number).