Is there a general formalism for 3-variable quadratic diophantine equations without mixed terms? Consider a polynomial equation
$$P_x(x) + P_y(y) + P_z(z) = 0\tag1 $$
Where $P_x, P_y, P_z$ are polynomials of degree at most two with integer coefficients.
The problem is to characterize all integer triplets $(x,y,z)$ satisfying $(1)$.
By "characterize" I mean to find some expressions in $n$ integer parameters to use as $(x,y,z)$, perhaps with restrictions on those integer parameters; or to be able to say that there is some finite set of solutions; or to say no solutions exist.
For example, the polynomials $P_x(x) = x^2$, $p_y(y) = y^2$, $P_z(z) = -z^2$ is the familiar Pythagorean equation and has a three-parameter family of solutions 
$$x = 2krs, y = k(r^2-s^2), z = k(r^2+s^2)$$ with $\gcd(r,s)=1$ and one of $r,s$ even; it also has one-parameter families $x = k, y= 0, z = \pm k$, and of course by symmetry it has solutions switching $x$ and $y$.
Without loss of generality, we can restrict the form of our polynomials to contain no linear terms (since we could always shift a variable and perhaps scale by two to get another equation without the linear term) and require that only one constant term appears. Thus I am asking:

Given, 
  $$ c_x x^2 + c_y y^2 + c_z z^2 = c_0 \tag2$$
  is there any general formalism for attacking $(2)$ ?

For specific $(c_x, c_y, c_z, c_0)$ it is usually possible by clever manipulations to solve the problem.  Pell's equation, for example, is a special case (with $c_x = c_0 = 1$ and $c_z = 0$).  
But is there a general methodology, and do any of these equations $(2)$ have only one or a finite number of solutions?
 A: You are asking about the integral automorphism group of an indefinite (ternary) quadratic form. Suppose we take $A,B,C> 0$ and ask about
$$ A x^2 - B y^2 - C z^2.  $$ 
We therefore have a symmetric matrix
$$ M =
\left(
\begin{array}{ccc}
A & 0 & 0 \\
0 & -B & 0 \\
0 & 0 & -C
\end{array}
\right).
 $$ 
A member of the (proper) automorphism group is a three by three matrix of integers $P$ such that 
$$ \color{magenta}{ P^T M P = M}.   $$
Next, taking $(x,y,z)$ as a row vector and $(x,y,z)^T$ as a column vector, we have the value of the quadratic form as 
$$  (x,y,z) M (x,y,z)^T  $$
So, here is the thing. When $P^T M P = M,$ then
$$ (P^T (x,y,z)) \; M (P (x,y,z)^T)   $$
comes out the same. 
And the integral automorphism group is infinite, and typically quite messy to work out in full detail.
It is a theorem of C.L. Siegel that the set of solutions to $Ax^2 - B y^2 - C z^2 = F$ occur in a finite number of distinct orbits under the action of the automorphism group. 
Meanwhile, there are usually two infinite subgroups that are easier to deal with. If $AB$ is not a perfect square, then $Ax^2 - B y^2$ has its own infinite automorphism group, which alters $(x,y)$ but leaves $z$ alone. If $AC$ is not a perfect square, then $Ax^2 - C z^2$ has its own infinite automorphism group, which alters $(x,z)$ but leaves $y$ alone. 
If you think it, ummm, unattractive, that the full group should not have a parametrization, see https://mathoverflow.net/questions/141284/integral-orthogonal-group-for-indefinite-ternary-quadratic-form
A: If $c_x, c_y, c_z, c_0$ all have the same sign, there are only finitely many solutions, which satisfy $|x| \le \sqrt{c_0/c_x}$ etc. (so a search will find them).
If $c_x, c_y, c_z $ have one sign and $c_0$ has the opposite sign, there are no solutions.
Since the set of solutions is invariant under sign changes, the only cases where there is a unique solution are those where the only solution is $(0,0,0)$.
A: Given an initial solution $u,v,w$ to,
$$au^2+bv^2+cw^2 = d$$
then an infinite more can be found using the identity,
$$a x_1^2 + b x_2^2 + c x_3^2 = d(x^2+bcy^2)^2\tag1$$
where,
$$\begin{aligned}
x_1\,&=u x^2 + b c u y^2\\
x_2\,&=v x^2 - 2c w x y - b c v y^2\\ 
x_3\,&=w x^2 + 2b v x y - b c w y^2
\end{aligned}$$
provided you choose $x,y$ such that,
$$x^2+bcy^2 =\pm 1\tag2$$
Thus, if the product of any two of $a,b,c$ is negative, then re-label so you can use the Pell equation $(2)$.
