3a^2 + 3a + 1 = n^2

If you multiply this equation by 4 and subtract 1, you can recast it as:

3(2a+1)^2 = (2n+1)(2n-1)

3 can factor only 1 of the factors of the RHS, which gives us 2 cases:

1) 2n+1 = p^2

2n-1 = 3q^2

2) 2n+1 = 3p^2

2n-1 = q^2

1) implies 3q^2 + 2 = p^2, which is impossible as no square = 2mod3.

So 2n-1 = q^2, which answers the problem.

Going further and setting q=2r+1 gives n = r^2 + (r+1)^2.

So you have the neat conclusion that when the difference of 2 consecutive cubes is a square, the root of that square is the sum of 2 consecutive squares.