Integer solutions of $x^3 = 7y^3 + 6 y^2+2 y$?

Does the equation $$x^3 = 7y^3 + 6 y^2+2 y\tag{1}$$ have any positive integer solutions? This is equivalent to a conjecture about OEIS sequence A245624.

Maple tells me this is a curve of genus $1$, and its Weierstrass form is $s^3 + t^2 + 20 = 0$, with $$ \eqalign{ s = \dfrac{-2(7 y^2 + 6 y + 2)}{x^2},& \
t = \dfrac{-2(3 x^3 + 14 y^2 + 12 y + 4)}{x^3}\cr
x = \dfrac{-2s(t-6)}{s^3+56},&\ y = \dfrac{4t-24}{s^3+56}}$$
So I can find rational points on both curves, but I haven’t been able to find integer points on (1) other than the trivial $(0,0)$.

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$$ \gcd(y, 7y^2 + 6 y + 2) = 1,2 $$

The first case is odd $y,$ so that $7y^2 + 6y+2$ is odd and $\gcd(y, 7y^2 + 6 y + 2) = 1.$ Both $y$ and $7y^2 + 6 y + 2$ must be cubes. Take $y = n^3.$ We want $7n^6 + 6 n^3 + 2$ to be a cube. Cubes are $1,0,-1 \pmod 9.$ If $n \equiv 0 \pmod 3,$ then $7n^6 + 6 n^3 + 2 \equiv 2 \pmod 9$ and is not a cube. If $n^3 \equiv 1 \pmod 9,$ then $7n^6 + 6 n^3 + 2 \equiv 6 \pmod 9$ and is not a cube. If $n^3 \equiv -1 \pmod 9,$ then $7n^6 + 6 n^3 + 2 \equiv 3 \pmod 9$ and is not a cube.

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$\gcd(y, 7y^2 + 6 y + 2) = 2.$ Both $y= 4n^3$ and $7y^2 + 6 y + 2 = 2 w^3$ .
give me a minute, it is not guaranteed to be easy just because the other case was. Hmmm, it is possible both mod 9 and mod 7, which reflects the solution with my $n=0, w=0.$ Sigh. Just taking $y=4u,$ there may be a tractable way to deal with $56u^2 + 12 u + 1 = w^3.$

Monday: computer suggests the only integer point on $56u^2 + 12 u + 1 = w^3$ is $(0,1),$ which would finish the problem if confirmed. CONFIRMED: see Integral solutions to $56u^2 + 12 u + 1 = w^3$

For what it’s worth, I took a generator $P$ of the Mordell-Weil group of $y^2=x^3-20$ and computed the preimage of $k\cdot P$ for $|k| \leq 300$ back to the original curve, and the only instance where the point was integral was for $k=0$.