A discussion about exponential diophantine equations and pythagorean triples 
Do all pythagorean triples $(a,b,c)$ have the identity that the three
(exponential diophantine) equations
\begin{equation} a^x+b^y=c^z \end{equation} \begin{equation}
 b^y+c^z=a^x \end{equation} \begin{equation} a^x+c^z=b^y \end{equation}
all have no more than one positive integer solution $(x,y,z)$?

This question is inspired by three exponential diophantine equations (1)$3^x+4^y=5^z$,(2) $4^y+5^z=3^x$, and (3) $3^x+5^z=4^y$.
Equations (1)(2)(3) all have no more than $1$ unique integer solution. Here, I'll give a proof for equation (3).


Find all positive integer solutions $(x,y,z)$ of the equation $3^x+5^z=4^y$.

$\because 3^x+5^z=4^y$, $\therefore (-1)^x+1^z\equiv 0\;\; (\operatorname{mod} 4)$. Also, since $x,y,z$ are all positive integers, so $x$ must be a multiple of $2$. Let $x=2a$. Then $3^{2a}+5^z=2^{2y}$, so we can get $$5^z=\left(2^y+3^a\right)\left(2^y-3^a\right).$$
Suppose $2^y+3^a=5^m, 2^y-3^a=5^n\;\;(m,n\in\mathbb{N}, m>n)$, subtracting the two equations, we get $2\cdot3^a=5^m-5^n$. If $n\geqslant1$, then $5\mid 5^m-5^n$, so $5\mid 2\cdot3^a$, which is a contradiction. Therefore, $n=0$.
We have $2^y=3^a+1$. For $a=1$, we get $y=2$. For  $a\geqslant 2$, writing the equation in  modulo $4$, we get $0\equiv (-1)^a+1$, so $a$ must be odd. Let $a=2k+1$, then $3^{2k+1}+1=2^a$. i.e. $3\cdot 9^k+1=2^y$. Writing the equation in modulo $8$, we get $3\cdot 1^k+1\equiv 0$. Which is another contradiction, so $(a,y)=(1,2)$, which leads us to $x=2a=2$.
Plugging in $x=2, y=2$ into the original equation (equation (3)), we get $5^z=7$. Therefore, $z$ is not an integer, so the exponential
diophantine equation $3^x+5^z=4^y$ doesn't have any solutions.

I'm curious that is the above statement (at the beginning of the question) true for all pythagorean triples. e.g. Does the exponential
diophantine equation $20^x+29^z=21^y$ have no more than one unique solution? Is there a general way to tackle with this problem?
Note:

*

*Again, I hope my question is not a duplication.

*$\mathbb{N}$ is the set of positive integers.


An argument mod $3$ shows that $z$ is even then let $z=2{z_1}.$
Hence   $3^x =  2^{2y} - 5^{2z_1}  = ( 2^y + 5^{z_1})(2^y - 5^{z_1})$
$\bullet\ 2^y - 5^{z_1}\ne 1$
$2^y + 5^{z_1} \pmod 3 \implies z_1 \not\equiv y \pmod 3$
$2^y - 5^{z_1} \pmod 3 \implies z_1 \equiv y \pmod 3$
This is a contradiction.
$\bullet\ 2^y - 5^{z_1} = 1$
According to Catalan's conjecture, $2^y - 5^{z_1} = 1$ has no positive integer solutions.
Hence $3^x+5^z=4^y$ has no positive integer solutions.
