you are simply being obstinate. You keep substituting numbers into Beal's equation. The expression cannot be factorised without numbers.
That's not what I did at all. In fact, the numbers I substituted wouldn't be valid to sub into Beal's equation (I made sure of this). It seems you have a problem with the distiction between factorising something algebraically and a prime factorisation.
Remember that every positive number greater than one has a unique prime factorisation, but not every polynomial (or other function) can be factorised nicely in the way you are thinking. Does this mean that once numbers are substituted the result will have no prime factors? No.
Remember that you are trying to prove that if we have two products added together and they equal another number (which of course they must) then all three numbers must have a common factor. This is clearly not true, as shown by a number of posters. To show this, simply add together any
two numbers with highest common factor 1. You clearly get another number which must have a prime factorisation and the original two numbers have no common factor. This contradicts what you are saying in your 'simple proof'.
I am aware I am probably wasting my time trying to convince you.