Understanding a proof that, if $xy$ divides $x^2+y^2+1$ for positive integers $x$ and $y$, then $x^2+y^2+1=3xy$ This is a Worked Example from Brilliant.org's entry on Vieta Root Jumping.

Let $x$ and $y$ be positive integers such that $xy$ divides $x^2+y^2+1$. Prove that $$x^2+y^2+1=3xy$$

The solution proposes that $x^2+y^2+1=k(xy)$ where $k$ is an integer.
It claims that there exists a minimum solution $(x,y)$ that has the minimum value of $x+y$.
So, they use $t$ to replace $x$ to show that $t^2-kty+y^2+1=0$
Then $t_1=x$ is one solution. By vieta formula,$t_1+t_2=ky$
Then $t_2=ky-x=\frac{x^2+y^2+1}{x}-x=\frac{y^2+1}{x}$
which implies $t_2\lt y$ then $t_1+t_2\lt x+y$.
So, the minimum condition only exists when $x=y$
I am ok as far, but after that it says, $x^2$ divided by $2x^2+1$, $x^2$divided by $1$.
So $k=3$.

But, why they can get $k=3$? $k=3$ only when $x$ and $y$ be the minimum solution. Why $k$ cannot be multiple of $3$? 


The proof you've given is incomplete.
You should start assuming $k \neq 3$, so you get $x \neq y$ (otherwise $k=3$ which is absurd).
Then, we have
$x \geq y+1$,
$x^2 \geq (y+1)^2 > y^2+1$,
$x>\frac{y^2+1}{x}=t_2$.
So $(t_2, y)$ is another solution to $\frac{x^2+y^2+1}{xy}=k$ with $k \neq 3$. But $t_2+y<x+y$, which contradicts the minimality condition.
