Solving $X^2-6Y^2=Z^3$ in positive integers I’m trying to solve the Diophantine equation
$$X^2-6Y^2=Z^3  \tag{$\star$}$$
in positive integers $x,y,z$.
Brute force calculations confirm the naïve intuition that there are many [read: surely infinite!] primitive solutions; numerical observation suggests the solutions have determinable [perhaps even “descent”- or “recurrence”-based?] patterns. For example, one “stream” of solutions is
$$(x,y,z) \in \{(5,2,1),\ (49,20,1),\ (485,198,1),\ (4801,1960,1),\dots\},$$
where $z=1$ and the $(x_n,y_n)$  satisfy the recurrence $t_n = 10t_{n-1}-t_{n-2}$. Another such stream exists for $z=19$, another for $z=25$, etc.
Evidently, ($\star$) is related to [but different from] the Pellian equation
$$X^2-6Y^2=Z^2,$$
so I’m wondering:

Question #1: Is a complete solution already known, either for the equation or for one of the variables (e.g. characterization of $z$)?


Question #2: If not, what are the most promising ways to attack the problem?

 A: For fixed $Z$ you have a Pell-type equation in $X$ and $Y$; by the theory of such equations, if there is one solution to this there are infinitely many.  In fact, if $(X,Y,Z)$ is one solution then $(5X+12Y,2X+5Y,Z)$ is another. The real question is for what values of $Z$ are there solutions?
EDIT: It appears that $X^2 - 6 Y^2 = Z^3$ has integer solutions if and only if $X^2 - 6 Y^2 = Z$ has solutions.  One direction is easy: if $X^2-6 Y^2 = Z$ then $(XZ)^2 - 6 (YZ)^2 = Z^3$.
EDIT: The nonnegative integers $z$ for which $X^2 - 6 Y^2 = z$ has integer solutions are OEIS sequence A242661.
On the other hand, you don't always have solutions with $\gcd(X,Y,z)=1$.  The sequence of $z$ for which $X^2-6Y^2=z^3$ has such solutions is
$1, 19, 25, 43, 67, 73, 97, 115, 139, 145, 163, 193, 211, 235, 241, 265, 283, \ldots$, which is not (yet) in OEIS.
A: This answer does not solve the problem.  It only makes an observation at the elementary level.  The constructed solution set is only a subset of the countable infinite solution set.

Let,
$$X=Ym, ~Z=Yn$$
$$\begin{align}&\implies Y^2(m^2-6)=Y^3n^3 \\
&\implies m^2-6=Yn^3 \\
&\implies Y=\frac{m^2-6}{n^3}\end{align}$$
If $n=1,~m≥3$, then
$$X=m(m^2-6),~Y=m^2-6, \\
Z=m^2-6 $$
$$X=m^3-6m,~ Y=m^2-6, \\
Z=m^2-6.$$
A: Setting (with Euler):
$x+y\sqrt{6}=(p+q\sqrt{6})^{3}$,
We get:
$x=p^{3}+18p.q^{2}$
$y=3p^{2}q+6q^{3}$
therefore
$z=p^{2}-6q^{2}$
with $(p,q)∈N$ and $p>q\sqrt{6}$
Ex.:
$(p,q)=(3,1)$,
$(x,y,z)=(81,33,3)$;
$(p,q)=(4,1)$,
$(x,y,z)=(136,54,10)$;
$(p,q)=(5,1)$,
$(x,y,z)=(215,81,19)$;
$(p,q)=(5,2)$,
$(x,y,z)=(485,198,1)$;
$(p,q)=( 2450,1000)$,
$(x,y,z)=( 58806125000, 24007500000,2500)$.
A: We can always construct a linear relation between x and y such as $x=ay+b; a, b \in \mathbb z$ to convert pell  equations  $x^2-Dy^2=1$ to a single unknown equation. We can find numerous families of solutions by this method. For Pell like equations we use rational solutions. I try to show this bellow by an example:
Let $x=y+a$ we have:
$x^2-6y^2=1\Rightarrow 5y^2-2ay-a^2+1=0$
$\Delta'=6a^2-5$
which have number of solutions:
$(a, \Delta')=(1, 1), (3, 7^2), (7, 17^2), \cdot \cdot\cdot$
which gives:
$(x, y)=(\frac 25, 1), (\frac 4 5, 1), (8, 5), (\frac {11}5, -\frac 45)\cdot\cdot\cdot $
For example take $(x, y)=(\frac {11}5, -\frac 45)$ we have:
$\big(\frac {11}5\big)^2-6\big(\frac 45\big)^2=1\Rightarrow\big(\frac {11^2}{25}\big)-6\big(\frac{4^2} {25}\big)=1$
We can rewrit this relation as:
$(11\times 25)^2-6(4\times 25)^2=25^3$
So: $(x, y, z)=(675, 100, 25) $
Similarly we can find more solutions where $z=25$ as a family of solutions.
Generally there can alway be a solution for equation $x^2-Dy^2=z^{2k+1}$ . for example  for $x^2-6x^2=z^5$ we can write:
$(11\times625)^2-6(4\times 625)^2=25^5$
For D=19 we have:
$x=\frac{35} 3$ , $y=\frac 83$ and we have:
$(35\times 9)^2-19(6\times 9)^2=9^3$
$(35\times 81)^2-19(8\times 81)^2=9^5$
A: I’m taking individ’s comment and spinning it out into an answer, in part to invite others to help parse it.
The formulas given on this AoPS page do indeed yield an identity solving ($\star$) in my original question.
There is some simplification which can be done in order to clarify the formulas. Writing $b=r-s$, and “compressing” Pell equations [wherever I found them, anyway!] yields
\begin{align}
  x &= \bigl(p^2+qr^2\bigr)\bigl(p^2(p^2+3qr(r-2s))-q(3p^2+qr(r-2s))(r-2s)^2\bigr),  \\[0.5em]
  y &= 2p\bigl(p^2+qr^2\bigr)\bigl(p^2(r-3s)+q(r+s)(r-2s)^2\bigr),  \\[0.5em]
  z &= \bigl(p^2+qr^2\bigr)\bigl(p^2+q(r-2s)^2\bigr).
\end{align}
Substitution into ($\star$) verifies that these formulas hold identically.
Evidently, solving $p^2+qr^2=\pm 1$ — always possible when $q$ is a nonsquare negative number — will eliminate the [algebraic] common factor $p^2+qr^2$; of course, using the minus sign will introduce a sign change in $z^3$ [but not $x^2$ or $y^2$]; furthermore, if one can simultaneously solve $p^2+q(r-2s)^2=\pm 1$ [using the same sign as the first choice], then we recover $z=1$.
I’m making this a community wiki answer, so that others will contribute. I’m hoping we can (a) find optimal formulas for $x,y,z$; (b) prove or disprove that this is a complete solution; and (c) perhaps determine how to use this technique [which individ appears unwilling to illuminate] to find similar “parameterizations” for other Diophantine equations.
I am writing $p^2 - 6 q^2 = r^3.$  It is important to allow the variables to be positive or negative in recipe B, as needed. Note that we may take absolute values of $p,q,r$ without changing anything. That allows the natural sequences, Fibonacci  type. For example $r=1$  uses the sequence of (positive) $p$ values $1, 5, 49, 485, 4801, $   made with rule $r_{k+2}  = 10 r_{k+1} - r_k$  Ummm. Taking $x,y$ coprime and possible conditions mod(2,3) will cause $\gcd(p,q,r) = | u^2 - 6 v^2 |$
Progress: Getting three recipes by the original, then multiplication by units $5 + 2 \sqrt 6$  or by $49 + 20 \sqrt 6,$  appears to allow us to use only $x,y \geq 0$. There is considerable repetition. The concept is that: we demand $x,y$ coprime; then $u,v$ coprime; and finally $w$, if larger than one, a prime $\equiv 7,11,13,17 \pmod{24}$ or the product of such primes.
Recipe A:
\begin{align}
p &= w^3 (x^3 + 18xy^2)  \\
q &= w^3 (3x^2y + 6y^3)  \\ 
r &= w^2 (x^2 - 6y^2)
\end{align}
Recipe B:
\begin{align}
p &= w^3 \color{blue}{(5x^3 + 36x^2y + 90xy^2 + 72y^3)}  \\
q &= w^3 \color{green}{(2x^3 + 15x^2y + 36xy^2 + 30y^3)}  \\
r &= w^2 (x^2 - 6y^2)
\end{align}
