So, a while ago, I watched a YouTube video about the positive integer solution of $$3^x + 4^y = 5^z$$ and the result was $ x = y = z = 2 $.
My question, now, is:
For any primitive Pythagorean triples $x, y, z$, does $x^a + y^b = z^c$ have any positive integer solution, other than $a = b = c = 2$?
And, does the following conjecture holds?
For any primitive Pythagorean triples $x_n, y_n, z_n$, make a set $P_n$, consisting of $x, y,$ and $z$, such that $\min(P_1)<\min(P_2)<\min(P_3)<\min(P_4)$ and so on. Then, the number of positive integer solutions of $x_k^a+y_k^b=z_k^c$, with $x_k, y_k, z_k$ a member of the set $P_k$, is less than $k.$
I tried attempting the first question, with $(5, 12, 13)$, but I haven't tried further triples yet. (The answer was, still, $x = y = z = 2$).
Andrew Wiles proved that $A^x+B^x=C^x$ is true only for $x=2$ but, if you allow different exponents, I can think of $$(1^1+2^3=3^2),\quad (9^1+3^3=6^2),\quad (17^1+4^3=9^2),\\ (19^1+5^3=12^2),\quad (40^1,6^3,16^2),\quad (18^1,7^3,19^2)$$
There are an infinite number of these triples and you can find them by subtracting any cube $(B)$ from the next higher square $(C)$ to get $(A)$
There are other easy ones like $(2^1 +5^2 +3^3).\space $ A $6$-deep For-loop finds many more like $(3^5 +10^2 =7^3),\space (7^3 +13^2 =2^9)$.
I'm just an amateur so I do not understand your conjecture.
Here is a simple program that may help you explore these latter triples.