A square number and a positive cubic number which differ by six Is there a square number and a positive cubic number (both positive integers) which differ by six?  If not, how do we prove this?

Here is a proof that $y^2 = x^3 - 6$ has no solutions, taken from here.
First note modulo $8$ that $y^2$ is one of $0,1, 4$ and $x^3$ is one of
$0, 1, 3, 5, 7$.
Thus, we must have $y^2 \equiv 1$ and $x^3 \equiv 7$.
I.e., $y$ is odd and $x \equiv 7 \pmod 8$.
Now write
$$
y^2 - 2 = (x - 2)(x^2 + 2x + 4)
$$
Since $x - 2 \equiv 5 \pmod 8$, a prime $p \; \mid \; x - 2$ congruent to $3$ or $5$ mod $8$.
Then $p \; \mid \; y^2 - 2$,
so $2$ is a perfect square modulo $p$.
But by a property of the Legendre symbol,
letting $p = 8k \pm 3$, we have
$$
\left( \frac{2}{p} \right)
= (-1)^{\frac{p^2 - 1}{8}}
= (-1)^{8k^2 \pm 6k + 1} = -1,
$$
which is a contradiction.
