Solution 1:

Let $g:\mathbb R^2\to\mathbb R$ be the map $(a,b)\mapsto a+b$. Clearly for any $t\in\mathbb R$, $$g^{-1}((-\infty,t])=\{(a,b)\in\mathbb R^{2}:a+b\leqslant t\}$$ is a measurable set. Therefore $g$ is a measurable map, so $X+Y=g(X,Y)$ and $\sigma(g(X,Y))\subset \sigma(X,Y)$. It follows that if $E\in\sigma(g(X,Y))$, $F\in\sigma(Z)$, then $E\in\sigma(X,Y)$ so that $$\mathbb P(E\cap F)=\mathbb P(E)\mathbb P(F),$$ from which we conclude.