When is $c^4-72b^2c^2+320b^3c-432b^4$ a positive square? In trying to solve a certain [third-degree] Diophantine equation, I have used the quadratic equation to determine that
$$c^4-72b^2c^2+320b^3c-432b^4$$
must be a positive integer square, where $c$ and $b$ are positive integers. 
From [admittedly limited] maxima calculations, it appears that the complete solution set is $(c,b) = (27k,5k)$ for $k = 1,2,3,\dots$.
I've tried to prove this, but have had no luck so far. Any hints or links/references to useful papers and techniques would be greatly appreciated.
EDIT: The original equation I was trying to solve is
$$y^2 = x^3 - 2. \tag{$\star$}$$
I am aware of the solution to this equation (see, for example, Conrad's paper, and about a dozen questions on this site), and so am not looking for a direct solution to ($\star$). Rather, I am interested in discovering techniques — especially elementary ones — by which one may attack the [type of] equation in the question.
 A: In general, an equation,
$$F(x) = y^2$$
with an initial rational point $x$ and where $F(x)$ is a quartic polynomial can be transformed into an elliptic curve, hence there should be an infinite number of rational points $x$. It is the case for your particular curve,
$$x^4 - 72 x^2 + 320x - 432 = y^2$$
with,
$$x_1 = c/b = 27/5$$
$$x_2 = c/b = 649193045949/48488608130$$
where $x_2$ is not reducible to $x_1$ and so on, ad infinitum.
(P.S. I've probably missed a smaller solution.)

$\color{blue}{Edit:}$ Here's an elementary method how to find more solutions $x_i$. 

(I believe it is called the tangent method and goes back to Fermat.) Simply equate,
$$x^4 - 72 x^2 + 320x - 432  =(a_2x^2+a_1x+a_0)^2$$
collect powers of $x$ to get,
$$p_4x^4+p_3x^3+p_2x^2+p_1x+p_0 = 0\tag1$$
and for this particular quartic, your three unknowns $a_0,a_1,a_2$ are enough to set $p_4 = p_3 = p_2 = 0$ and making the $x^4, x^3, x^2$ vanish, leaving you with,
$$x = -p_0/p_1 = 27/5$$
To get more solutions, do the substitution $x = v+27/5$, to get,
$$v^4+108/5v^3+2574v^2/25+21532/125v+(171/25)^2 = (b_2v^2+b_1v+b_0)^2$$
which you assume to be a square in the RHS. Collect powers of $v$,
$$q_4v^4+q_3v^3+q_2v^2+q_1v+q_0 = 0\tag2$$
Your three unknowns $b_0,b_1,b_2$ are also enough to set the other end $q_0 = q_1 = q_2 = 0$ and making the $v^0, v^1, v^2$ vanish, leaving you with,
$$q_4v^4+q_3v^3 = (q_4v+q_3)v^3  = 0$$
Hence,
$$x_2 = v+x_1 = -q_3/q_4+27/5 = 649193045949/48488608130$$
I'm sure with this easy method, you now have a technique to generate new solutions from old ones.
