Beal's conjecture states: If $A^x + B^y = C^z$ for $A, B, C, x, y ,z$ are positive integers and $x,y,z>2$ then $A$ , $B$ and $C$ have a common prime factor.
As I understand, if for example $x$ is even, we can write $A^x=(A^{x/2})^2$ and $A^{x/2}$ is integer. We can do similar for factors of $x$ and keep one prime as the exponent. A similar approach is usually used when Fermat's Last Theorem is discussed.
Question: is it so that the conjecture allows only odd exponents? and is not it enough to prove the conjecture for only odd prime exponents (again similar to FLT discussions)?
If the answer to this question is yes, then why Wikipedia's page of Beal's conjecture presents special cases where one of the exponents is $2$ or another even number? For example why case $(x, y, z) = (3, 6, n)$ is permitted in Beal's conjecture?
Similar cases have been presented in many websites, such as here where example $27^4 + 162^3 = 9^7$ is presented. It has been mentioned that this is not a counterexample because the base numbers share a factor, but isn't it that this is not even permitted in the conjecture?
Moreover, is there any example that fits in the equation of Beal's conjecture, the exponents are larger than 3 and the base numbers share a factor?