Infinitely many solutions to $a^x+b^y=c^z$ for given $ x, y, z \in \mathbb{N}$? Problem I have recently seen a few questions such as this(1) and this(2) and this has led me to wonder what conditions are sufficient on $x, y, z \in \mathbb{N}$ so that when given $x, y, z$,  $a^x+b^y=c^z$ has infinitely many solutions. This seems like a well known problem, but I have not been able to come across it before so perhaps someone could enlighten me.
Progress:
I have already shown that it is sufficient (but by no means necessary) that $x, y, z$ are pairwise coprime.
Proof:
Though many proofs can be found, I will write one (rather nice) proof here.
We first notice that $$2^p3^q+2^{p+1}3^q=2^p3^q(1+2)\\ =2^p3^{q+1} \ \forall p, q \in \mathbb{N}$$
It would be great if $2^p3^q, 2^{p+1}3^{q} \ \mbox{and} \ 2^p3^{q+1}$ were powers of $x, y, z$ respectively. However, we can find sufficient conditions on $p, q$ so that such is the case. The following are the sufficient conditions:


*

*$p,q \equiv 0 \mod{x}$

*$p+1, q \equiv 0 \mod{y}$

*$p, q+1\equiv 0\mod{z}$
Then $2^p3^q$ has an exponent divisible by $x$, $2^{p+1}3^q$ has an exponent divisible by $y$, and $2^p3^{q+1}$ has an exponent divisible by $z.$
In other words, if we can find infinitely many $p$ and $q$ so that the three above conditions are satisfied, we have found infinitely many solutions to $a^x+b^y=c^z$ given $x, y, z \in \mathbb{N}$ pairwise coprime. But the conditions are equivalent to the following:
$$p\equiv \begin{cases}0 \  \ \ \mod{x}
\\ 0 \ \ \ \mod{z}
\\-1 \mod{y}
\end{cases}$$
and
$$q\equiv \begin{cases}0 \  \ \ \mod{x}
\\ 0 \ \ \ \mod{y}
\\-1 \mod{z}
\end{cases}$$
But since $x, y, z$ are pairwise coprime we can use the Chinese Remainder Theorem, giving us unique solutions for $p$ and $q$ modulo $xyz$. In particular, we have found infinitely many values of $p$ and $q$, which is what we wanted.
By no means are these conditions necessary, consider the simple case $a^2+b^2=c^2$ which is known to have also infinitely many solutions, but in this case $x, y, z$ are not pairwise coprime. So for what other $\langle x, y, z \rangle$ gives rise to infinitely many solutions to $a^x+b^y=c^z$?
 A: If $\gcd(z,xy)=1$, then by chinese reminder theorem; 
there is infinitly many $k$ such that: 
$$ 
k \overset{xy}{\equiv} 0 
\ \ \ \ 
\text{and} 
\ \ \ \ 
k \overset{z}{\equiv} -1; 
$$
now let 
$$
a:=2^{   \tiny{   \dfrac{k}  {x}    }   }, 
\ \ \  
\ \ \  
b:=2^{   \tiny{   \dfrac{k}  {y}    }   }, 
\ \ \  
\ \ \  
c:=2^{   \tiny{   \dfrac{k+1} {z}   }   }; 
$$
and notice that: $\ 2^k+2^k=2^{k+1}$.


Let's consider the equation: 
$$x^p+y^q=z^r;$$ 
with $2 \leq p, q, r$. 


*

*If $ \ \dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{r} \leq 1 $ ;
then there is no polynomial parametric solution for coprime $x,y,z$.  

*If $ \ 1 < \dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{r} $ ;
then there is polynomial parametric solution for coprime $x,y,z$.
This case contains the following cases:
$$ 
\ \ 
\ \ \ 
\ \ \ 
\ \ \ 
\{ p,q,r \} = \{ 2,3,6 \}, 
\ \ \  
\{ p,q,r \} = \{ 2,3,5 \}, 
\ \ \  
\\ 
\ \ 
\ \ \ 
\ \ \ 
\ \ \ 
\{ p,q,r \} = \{ 2,3,4 \}, 
\ \ \  
\{ p,q,r \} = \{ 2,3,3 \}; 
\ \ \  
\\ 
\{ p,q,r \} = \{ 2,2,n \}, 
\ \ \ 
\ \ \ 
\ \ \ 
\text{for every} 
\ 
n 
.  
$$ 
