Positive integer solutions of equations of the form $ x^4+bx^2y^2+dy^4=z^2$ A method used to answer the post Does the equation $y^2=3x^4-3x^2+1$ have an elementary solution? showed that the only positive integer solutions of the equations $$ x^4-3x^2y^2+3y^4=z^2$$ $$X^4+6X^2Y^2-3Y^4=Z^2,$$ are  $(1,1,1)$ and $(1,1,2)$, respectively. Furthermore these solutions are related in that each can be deduced from the other.
In general, let $b,d,D$ be integers such that $b^2=4d+D$ and consider the pair of equations  $$ x^4+bx^2y^2+dy^4=z^2\tag 1$$ $$X^4-2bX^2Y^2+DY^4=Z^2\tag 2$$ where we can assume that $(x,y)=(X,Y)=1$. Then for the method of the aforementioned post to produce an infinite sequence of solutions, we require a solution $(x,y,z)$ of $(1)$ to generate the solution $(\frac{z}{x},y,|\frac{x^4-dy^4}{x^2}|)$ of $(2)$ and we require a solution $(X,Y,Z)$ of $(2)$ to generate the solution $(\frac{Z}{2X},Y,|\frac{X^4-DY^4}{4X^2}|)$ of $(1)$.
I can prove that for this to occur $y=Y$ is odd. Also, since $x$ must be a factor of $z$, $x^2$ is a factor of $d$. Therefore there are only a finite number of possibilities for $x$ and so we are dealing with a finite loop of solutions. Beyond that, I only know that when solving specific equations these loops are rare and, when they occur, seem to have period $2$.
Questions
What periods are possible for such loops of solutions?
What else can be determined about them?

SOME PROGRESS
Let the loop of solutions have period $2n$. We can suppose that successive values taken by $x$ and $X$ are $a_1,a_2,...a_n$ and $A_1,A_2,...A_n$, respectively, where for each $i$ there are integers $c_i$ and $C_i$ such that $d=a_i^2c_i$ and $D=A_i^2C_i$.
Then the conditions given in the post simplify to:-
$$a_i^2+by^2+c_iy^4=A_i^2 \tag 1$$
$$A_i^2-2by^2+C_iy^4=4a_{i+1}^2 \tag 2$$
$$2a_{i+1}A_i=|a_i^2-c_iy^4| \tag 3$$
$$4a_{i+1}A_{i+1}=|A_i^2-C_iy^4| \tag 4$$
$y^2$ is a factor of $4^n-1$
Consider equations $(1)$ and $(2)$ modulo $y^2$.
$$a_1^2\equiv A_1^2\equiv 4a_2^2\equiv 4A_2^2\equiv 16a_3^2\equiv 16A_3^2\equiv ... $$ Hence $a_1^2\equiv 4^na_1^2$ and so, since $x$ and $y$ are coprime, $y^2$ is a factor of $4^n-1.$
This significantly limits the values of $y$. For periods of less than or equal to $18$ the only possible values for $y$ are $1$ and $3$.
A general formula for loops of period $2$
If $n=1$, then $y=1$ and the equations further simplify to
$$a^2+b+c=A^2 \tag 1$$
$$A^2-2b+C=4a^2 \tag 2$$
$$2aA=|a^2-c| \tag 3$$
$$4aA=|A^2-C| \tag 4$$
From equation $(3)$ we have $c=a^2\pm 2aA$. Then $b=A^2\mp 2aA-2a^2$ and $C=A^2\mp4aA.$
The original equations are then $$ x^4+\left(A^2\mp 2aA-2a^2\right)x^2y^2+\left(a^4\pm2a^3A\right)y^4=a^2A^2$$ $$X^4-2\left(A^2\mp 2aA-2a^2\right)X^2Y^2+\left(A^4\mp4aA^3\right)Y^4=4a^2A^2,$$
with a loop of solutions $(a,1,aA)$ and $(A,1,2aA).$
