Finding all integers $x$, $y$, $z$ satisfying $3x^2=y(y+2)=z^3-1$ 
Find all integers $x,y,z$ such that
$$3x^2=y(y+2)=z^3-1$$

This is one exercise in trial test of my local. First, I see that from $y(y+2)=z^3-1$, one obtains $(y+1)^2=z^3$. Therefore, $(y+1)^2=b^3$ and $z=c^2$.
This result is a direct consequence of the result which was prove by J. H. E. Cohn in https://sci-hub.se/https://doi.org/10.1093/qmath/42.1.27. But i wonder if we have any simplier way to approach?
Thanks
 A: Once you've made the observation that you made, you get $3x^2=s^6-1$ for some positive integer $s$. Factor this as
$$3x^2=(s^3-1)(s^3+1).$$
Now, the greatest common factor of the two factors on the right side is either $1$ or $2$. If it is $1$, one of the factors must be a square and the other must be $3$ times a square, and if it is $2$ then they must be $2$ and $6$ times squares (they can't be $1$ and $3$ times squares since one would be $\equiv 2\bmod 4$). This is a consequence of the property that, if a perfect square is factored into relatively prime factors, they are each squares.
To solve the first case, we examine the factor that is a pure square;
$$s^3\pm 1 = t^2.$$
This can be factored as $(s\pm 1)(s^2\mp s+1)=t^2$. The greatest common factor of these factors is either $1$ or $3$, but it can't be $3$, since $(s^3\pm 1)\mp 2$ is a multiple of $3$. Thus, $s^2\mp s+1$ is a square. However, this is strictly between $(s\mp 1)^2$ and $s^2$ for $s>1$, and for $s=1$ we have only the trivial $t=0\implies x=0$.
In the second case, we have
$$(s\pm 1)(s^2\mp s+1)=s^3\pm 1=2t^2.$$
Again, these two factors are relatively prime, so one is a square and one is twice a square. The square cannot be the second factor for the same reason as before, but that factor cannot be twice a square as it is odd. This concludes the proof.
