Let $t=x+y, r=|x-y|$, then $z=\frac{t^2-r^2}{t}\to v_2(t)=2v_2(r)>0$. Let $r=r_1*2^k, k=v_2(r), t=t_1*2^{2k}$, then $z=\frac{2^{2k}t_1^2-r_1^2}{t_1}$. Let $t_1=b^2c$, were $rad(c)=c$, then $r_1=abc$ and $z=(2^{2k}b^2-a^2)c$. Because $a$ is odd, $(2^kb)^2-a^2=3\mod 4$ and had prime divisor form $4m+3$.

Let $d=\gcd(x,y),x=ad,y=bd$, then $a$ and $b$ are coprime and $z=\frac{4dab}{a+b}$. Since $\gcd(ab,a+b)=1$, then $ab|z$. If $a$ or $b$ is of the form $4n-1$, then we are done. Otherwise $a+b\equiv2\pmod4$, so $\frac{4d}{a+b}$ is even, so $z$ is even, a contradiction.

We claim that $z$ is less than $x$ or $y$. For if $z>x,y$ then $\frac{4xy}{x+y}<x$, $\frac{4xy}{x+y}<y$, or $3x<y$, $3y<x$. But adding up gives $2(x+y)<0$, which is impossible. Therefore, we may assume (wlog) that $z=y+t$ for some $t\in \mathbb{Z}_+$. Substituting this into the equation and expanding yields $y^2+(t-3x)y+tx=0$, which by considering $\Delta_y$ is can be written as $(2y+t-3x)^2+(4x)^2=(5x-t)^2$. Hence there exist $r,s\in \mathbb{Z}_+$ such that: $2y+t-3x=r^2-s^2$, $4x=2rs$, $5x-t=r^2+s^2$; which after solving becomes: $x=\frac{rs}{2}$, $y=r^2-\frac{rs}{2}$, $z$ $=y+t$ $=2rs-s^2$; where one of $r,s$ is even and the other is odd (since $z$ must be odd). Now, if $r$ is even and $s$ is odd, then $z$ $=2rs-s^2$ $\equiv 3\pmod 4$ and the result follows. However, if $r$ odd and $s$ even then $z$ is even, which contradicts $z$ being odd.

Observe that $z=\frac{4dab}{a+b}$, where $d=(x,y), a=\frac{x}{d}, b=\frac{y}{d}$. Notice that $(ab,a+b)=1$, so that $4d=k(a+b)$ and $z=kab$. Since $z$ is odd, $k,a,b$ are also odd and $4|a+b$ due to the relation $4d=k(a+b)$. Therefore either $a$ or $b$ is of the form $4m-1$. It remains to observe that both $a$ and $b$ are divisors of $z$.