This question might be easy.
The hard question is this: prove that if $a,b,c\geq3$ then there are no solutions in positive integers $x,y,z$ to $x^a+y^b=z^c$ with $x,y,z$ coprime. This implies Fermat, most cases of Catalan, etc., and is an open problem.
But it's really crucial that $x,y,z$ are coprime to make this question hard. For example if I want to find any solution to $x^9+y^{10}=z^{11}$ in positive integers, I just start with a random solution to $A+B=C$, e.g. $1+1=2$, and now I multiply by an appropriate power of all the primes dividing $ABC$ to get a solution. For example, if I start with $1+1=2$ then I multiply both sides by $2^N$ and deduce $2^N+2^N=2^{N+1}$. Now it's easy to find a positive $N$ with $N=0$ mod 9, $N=0$ mod 10 and $N=-1$ mod 11, and for this value of $N$ we get a solution in positive integers to $x^9+y^{10}=z^{11}$.
But this trick relies on the fact that 9, 10, 11 are pairwise coprime. It wouldn't surprise me if an extension of the trick could give a solution in positive integers to $x^6+y^{10}=z^{15}$, where the point is that the exponents aren't pairwise coprime, but 5 minutes on the back of an envelope didn't give me the trick I needed, and I thought that here might be a great place to ask.
What's the trick I've missed?