Properties of three terms of a geometric series I’m [still!] working on the equation in this question, namely
$$(b^2+2)^2=(a^2+2c^2)(bc-a).  \tag{$\star$}$$
where $a,b,c$ are integers. Evidently, $(\star)$ implies
$$\frac{b^2+2}{bc-a} = \frac{a^2+2c^2}{b^2+2},  \tag{1}$$
which is to say that $\{bc-a,b^2+2,a^2+2c^2\}$ are three consecutive terms of a geometric series.

QUESTION: Does that fact provide any information that would help in solving $(\star)$? i.e., are there properties of geometric series that can be brought to bear on the problem?

Each fraction in $(1)$ is actually an integer, in case that provides more leverage/structure.
EDIT: The reason I know this is that I derived this equation from the equation $x^3=y^2+2$, where $x=(b^2+2)/(bc-a)$ is a positive integer by assumption.

If I have understood the additional constraints to your original problem correctly, the problem is now to find all integral solutions to
$$(b^2+2)^2=(a^2+2c^2)(bc-a),  \tag{$\star$}$$
such that
$$\frac{b^2+2}{bc-a}=\frac{a^2+2c^2}{b^2+2},$$
and moreover $x:=\frac{b^2+2}{bc-a}=\frac{a^2+2c^2}{b^2+2}$
is an integral solution to $x^3=y^2+2$ for some integer $y$.

A standard argument in $\Bbb{Z}[\sqrt{-2}]$, which is a UFD, shows that then $x=3$. Hence we can express the problem as a system of two simultaneous diophantine equations:
\begin{eqnarray}
a^2+2c^2&=&3(b^2+2)\tag{1.1}\\
b^2+2&=&3(bc-a).\tag{1.2}
\end{eqnarray}
Equation $(1.2)$ shows that $3a=3bc-b^2-2$, and hence from $(1.1)$ we find that
$$27(b^2+2)=9(a^2+2c^2)=(3a)^2+18c^2=(3bc-b^2-2)^2+18c^2.$$
Expanding and collecting like terms shows that this is equivalent to
$$b^4-6b^3c+9b^2c^2-23b^2-12bc+18c^2-50=0,$$
which in turn we can rewrite as
$$(b^2+2)(b-3c)^2=25(b^2+2).$$
Of course $b^2+2\neq0$ and so it follows that $(b-3c)^2=25$, or equivalently
$$b=3c\pm5\qquad\text{ and hence }\qquad a=\frac{3bc-b^2-2}{3}=\mp5c-9.$$
This shows that every solution to your system of diophantine equations is of the form
$$(a,b,c)=(\mp5c-9,3c\pm5,c).$$

Note that these are precisely the solutions you already found in your original question, i.e. the parametric family of solutions
$$(5d+1,3d+1,d+2)\qquad\text{ with }\qquad d\in\Bbb{Z},$$
along with their involutions given by $(a,b,c)\ \longmapsto\ (a,-b,-c)$.
