No, this is is hopeless, as becomes clear when you write
$$
\textrm{Cov}(X,Z)+\textrm{Cov}(Y,Z)=E(X+Y)Z . \tag{1}
$$
Your assumptions don't restrict $U=X+Y$ much, you only know that $EU=0$, $\textrm{Var}(U)=2$ and $U$ is [2-divisible.][1]

More to the point perhaps, you can just take $Z=(X+Y)/\sqrt{2}$ to obtain a counterexample, then (1) equals $\sqrt{2}$. This is also the largest possible value, by Cauchy-Schwarz.


  [1]: https://en.wikipedia.org/wiki/Infinite_divisibility_(probability)