My main question is: Is there a criteria for solving a quartic Diophantine equation of the form
$$x^4 + ax^3 + bx^2 + ax + d = 0 \tag 1$$
We have the restriction $d \ne 1$.
Here's my effort in obtaining a criteria and I would like the solution to be verified.
EDIT: As @mathlove points out in comments, this approach is incorrect. I am leaving the question as-is for other readers.
Eqn (1) is almost palindromic (or what Wikipedia calls as [Quasi-palindromic][1]). The suggested variable transformation for quartics of the form
$$a_0 x^4 + a_1 x^3 + a_2 x^2 + a_1mx + a_0 m^2 \tag 2$$
is to divide by $x^2$ and apply the change of variable $$z = x + \frac{m} {x} \tag 3$$ and obtain
$$ a_0 z^2 + a_1 z + a_2 - 2ma_0 = 0 \tag 4 $$
Using Eqns. (2),(3) and (4) and with $m=1$, our Eqn. (1) becomes
$$z^2 + az + b - 2d = 0 \tag 5$$
Using the quadratic formula
$$z = \frac {-a \pm \sqrt {a^2 - 4(b-2d)}} 2 \tag 6$$
This has real solutions iff the discriminant $$D = a^2 - 4b + 8d \gt 0. \tag 7$$
This is a necessary and sufficient condition for a real solution.
From the Integral Root Theorem, $x$, the root of Eqn. (1) must divide $d$.
So, the best I could come up with as a solvability criteria for Eqn. (1) is
$$ \begin{align} a^2 - 4b + 8d \gt 0 \land x \in \{d_i : \text{the set of divisors of $d$}\} \tag 8 \end{align} $$
Can this criteria be refined or is this the best we can get?