Let $r > s$ be two positive real numbers. Then, there would be a positive real number $t$ such that $t^2 + f(s)t + s - r = 0$. Now, put $(x, y) = (t, s)$ it follows that
$$f(r) = f(t^2 + f(s)t + s) = 2t + f(s) > f(s).$$
So, $f$ is strictly increasing and hence injective. Plugging $(x, y) = (\frac{f(z)}{2}, y)$ it follows that
$$f(z) + f(y) = f(f(z)^2/4 + f(z)f(y)/2 + y) = f(f(y)^2/4 + f(z)f(y)/2 + z)$$
That is, $f(z)^2/4 - z = C$. Hence, $f(z) = \sqrt{4z + C}$, for some $C \ge 0$. $\square$