Random Diophantine?

Setting $y = 0$ and $x = 1$ gives $2z + 1$ so all odd numbers are possible. Setting $(x, y) = (2, 0)$ gives $16 + 8z$ so all positive multiples of $8$ strictly greater than $8$ are possible. This gives a total of $1007 + 250 = 1257$ possible numbers.

Now, we show that there are no other numbers. Note that the expression can also be written in the form $(x^2 + y^2 + z)(x + y)(x - y)$ , so either all the factors are even or all of them are odd. This means that all non-multiples of $8$ and even numbers cannot be written in this form.

Finally, we show that $8$ is unattainable. Observe that $x > y$ so that $x + y = x - y = x^2 + y^2 + 2z = 2$ . This implies $y = 0, x = 2, z = -1$ , which is clearly impossible. Thus, there are only $\boxed{1257}$ possible numbers.