Suppose that $x,y$ are positive real numbers and that
$$ (x-y)^2=2\big( (x+y)-2\sqrt{xy} \big). \tag{*}$$
Then Mathematica claims that one of the following $3$ options holds:
$$1. \, \, \, x=y.$$ $$2. \, \, \, x = y - 2 \sqrt 2 \sqrt y + 2.$$ $$3. \, \, \, x = y + 2 \sqrt 2 \sqrt y + 2.$$
(Option $3$ is in fact impossible over the reals.)
Also, if I interpret the results correctly, the last two options $(2,3)$ are only possible when $xy \le \frac{1}{4}$. When $xy \ge \frac{1}{4}$ only option $(1)$ is possible.
Is there a way to prove this analytically (without using a computer)?
Edit:
We assume $x \ge 0,y \ge 0$. In the answer below, we rewrite equation $(*)$ as $(x - y)^2 = 2 ( \sqrt{x} - \sqrt{y})^2$, which gives
$$ (\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y})=x - y =\pm \sqrt{2} ( \sqrt{x} - \sqrt{y}).$$
Thus, either $x=y$ or $\sqrt{x} + \sqrt{y}=\pm \sqrt{2}$. If $x,y$ are non-negative reals, then $\sqrt{x} + \sqrt{y}=- \sqrt{2}$ is ruled out.
So, we are left with $$\sqrt{x} + \sqrt{y}= \sqrt{2} \Rightarrow \sqrt{x} = \sqrt{2}-\sqrt{y} \Rightarrow x=y - 2 \sqrt 2 \sqrt y + 2.$$
The third option actually comes from the branch where $\sqrt{x} + \sqrt{y}=- \sqrt{2}$, and then $x,y$ are complex numbers.
Next, we show that $\sqrt{x} + \sqrt{y}= \sqrt{2}$ is possible if and only if $xy \le \frac{1}{4}$.
By the AM-GM inequality $\frac{1}{\sqrt 2}=\frac{\sqrt{x} + \sqrt{y}}{2}\ge \sqrt{\sqrt{xy}}$.
On the other hand, suppose that $\sqrt{xy} =s \le \frac{1}{2}$. Writing $a=\sqrt x,b=\sqrt y$, we are looking for $a,b \ge 0$ such that $ab=s,a+b=\sqrt{2}$. This is a quadratic equation, and since the AM-GM holds it has real solutions, which must be positive.
(Indeed, since $s=ab$ is positive, then $a,b$ have the same sign, and $a+b=\sqrt{2}>0$ implies they are both positive.
The quadratic is $$ t^2-\sqrt 2 t+s=0.$$
Factorise to $(x - y)^2 = 2 ( \sqrt{x} - \sqrt{y})^2$. Which in turn gives $(\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y}) = \pm \sqrt{2} (\sqrt{x} - \sqrt{y})$. So $x=y$ or $\sqrt{x} = \pm \sqrt{2} - \sqrt{y}$.