If 2ˣ = 3ʸ = 6ᶻ, where $x,y$ and $z$ are all non-zero, which relation must hold?

Call the common value $k$. Then $2=k^{1/x},3=k^{1/y}$ and $6=k^{1/z}$. Because $2\times 3=6$, we get $k^{1/x}\times k^{1/y}=k^{1/z}$, and equal bases force equal indices: $1/x+1/y=1/z$. A numerical check with $k=36$ gives $x$ ≈ 5.17, $y$ ≈ 3.26 and $z=2$, and indeed 1/5.17 + 1/3.26 ≈ 0.5 = 1/2. Adding the indices themselves is the trap — it is their RECIPROCALS that add, because 6 is the product (not the sum) of 2 and 3.