If $a,b > 1$ and $r>2$ does $ax^2+by^2=z^r$ have any rational solutions?

I have been trying to solve the following equation for months without much success. It has been so far a very frustrating endeavor.Please help.

Consider the diophantine equation: $x^2+y^2=z^r$ where $\gcd(x,y,z)=1$ and $r>2$.
Assuming that there exists a non-trivial triplet: $(x_0,y_0,z_0)$ satisfying: $$x_0^2+y_0^2=z_0^r$$ How do I find the parametrization of:$$ax^2+by^2=z^r$$?

Solutions Collecting From Web of "If $a,b > 1$ and $r>2$ does $ax^2+by^2=z^r$ have any rational solutions?"

It’s quite easy to parameterize,

$$\color{red}ax^2+\color{red}by^2 =z^k$$

for odd $k$. Assume,

$$x^2+by^2 = (p^2+bq^2)^k$$

$$(x+y\sqrt{-b})(x-y\sqrt{-b}) = (p+q\sqrt{-b})^k(p-q\sqrt{-b})^k$$

Equate factors and solve for $x,y$. Hence,

$$x =\frac{\alpha+\beta}{2},\quad y = \frac{\alpha-\beta}{2\sqrt{-b}},\quad\text{where}\quad \alpha = (p+q\sqrt{-b})^k,\quad \beta = (p-q\sqrt{-b})^k$$

For example, let $k = 3$. Then,

$$(p (p^2 – 3 b q^2))^2 + b(3 p^2q – b q^3)^2 = (p^2+b q^2)^3$$

Let $p = \sqrt{a}p$. Then,

$$\color{red}a(ap^2 – 3 b q^2)^2 + \color{red}b(3 ap^2q – b q^3)^2 = (ap^2+b q^2)^3\tag{k=3}$$

for free variables $p,q$. Let $k = 5$,

$\color{red}a(a^2p^4 – 10 a b p^2q^2 + 5 b^2 q^4)^2 + \color{red}b(5 a^2p^4q – 10 a bp^2 q^3 + b^2 q^5)^2 = (ap^2+b q^2)^5\tag{k=5}$

and so on.