This is what I have so far:
$$x^3-x = 12y + 6$$
$$x(x+1)(x-1) = 2(6y+3)$$
The RHS of the equation is even, so therefore so must the LHS. Given that the three numbers on the LHS are consecutive,
then we know $x$ is even and $(x-1)$ and $(x+1)$ are odd.
Given that they're three consecutive numbers their gcd is 1, making them coprime.
That's as far as I got. Please note that both $x$ and $y$ are positive integers.
Note that necessarily $x\equiv2\pmod{4}$ because $$x(x-1)(x+1)=12y+6\equiv2\pmod{4},$$ and conversely if $x=4z+2$ then $$x^3-x=4(16z^3+24z^2+5z)+6,$$ where a quick check shows that $16z^3+24z^2+5z\equiv0\pmod{3}$ for all $z$.