In this part of the discussion, we will sometimes refer to $A,B,C$ instead of $x,y,z$ and we will look for triples where $A$ is an $even$ cube such that $A^2=x^3=C^2-B^2$. It is convenient to be able to find a triple for every pair of natural numbers and we can do so if we vary the standard formula to have an effect similar to $(m,n)=(2n+k-1,k)$. We can then find the desired triples with:
In these functions, $n$ is a set number and $k$ is the member number within that set. It is easy to find these triples in a spreadsheet where one column is dedicated to testing if the cube root of $A$ is an integer. Here are primitives ($A^2+B^2=C^2\land GCD(A,B,C)=1$) where each f(n,k) is a triple from sets $1$ thru $50$ and member numbers up to $300$.
We can do the same for side $A$ odd if we let A,B,C be:
There are an infinite number of these triples for side $A$ odd because the function for $A$ generates ever odd number $>1$. In a casual search, the primitives appear to be confined to $Set_1$ but this is not proven. Here are examples from $k=1$ to $k=2000$: