**Corrected after a comment of André Nicolas** :
We have $4^5=1024>1016$. This shows that $y$ can only be $1$, $2$ or $3$. Now if $y=3$, then since $3^2\times 3^5>1016$, the only possible values of $x$ are $1$ and $2$. Similarly, if $y=2$, the only possible values of $x$ are $1,2,3,4,5$. Computing the products $xy$ and comparing them with the values proposed in the question, out of these seven options there remain only two: $(x,y)=(2,2)$ and $(x,y)=(5,2)$. Checking for $z^3$, out of these two possibilities there remains only one: $(x,y,z)=(5,2,6)$.
Now if $y=1$, then $xy=x$, and all we have to check is: which of the six values among $1016-(xy)^2$ are cubes. This gives one more triple $(x,y,z)=(4,1,10)$.
So the possible values of $xy$ (among indicated) are $4$ and $10$.