Prove there is no $x, y \in \mathbb Z^+ \text{ satisfying } \frac{x}{y} +\frac{y+1}{x}=4$ Prove that there is no $x, y \in \mathbb Z^+$ satisfying
$$\frac{x}{y} +\frac{y+1}{x}=4$$ 
I solved it as follows but I seek better or quicker way: 
$\text{ Assume }x, y \in \mathbb Z^+\\  1+\frac{y+1}{y}+\frac{x}{y} +\frac{y+1}{x}=1+\frac{y+1}{y}+4 \\
\Rightarrow \left(1+\frac{x}{y}\right)\left(1+\frac{y+1}{x}\right)=6+\frac{1}{y}\\
\Rightarrow (x+y)(x+y+1)=x(6y+1)\\
\Rightarrow x\mid (x+y) \;\text{ or }\; x\mid (x+y+1)\\
\Rightarrow x\mid y \;\text{ or }\; x\mid (y+1)\\
\text{Put}\; y=nx ,n \in \mathbb Z^+\;\Rightarrow\; \frac{x}{nx} +\frac{nx+1}{x}=4 \;\Rightarrow\; \frac{1}{n} +\frac{1}{x}=4-n \;\rightarrow\;(1)\\
\text{But}\; \frac{1}{n} +\frac{1}{x} \gt 0 \;\Rightarrow\; 4-n \gt 0 \;\Rightarrow\; n \lt 4\\
\text{Also}\; \frac{1}{n},\frac{1}{x}\le 1 \;\Rightarrow\; \frac{1}{n} +\frac{1}{x} \le 2 \;\Rightarrow\; 4-n \le 2 \;\Rightarrow\; n \ge 2\\
\;\Rightarrow\; n=2 \;\text{ or }\; 3, \;\text{substituting in eq. (1), we find no integral values for } x.\\ \text{The same for the other case.}\\
$
So is there any other better or intelligent way to get this result?

For $x,y\in \Bbb Z^+$ we have  $$\frac {x}{y}+\frac {y+1}{x}=4\implies x^2-4xy+y^2+y=0\implies x=2y\pm \sqrt {3y^2-y}\implies$$  $$\implies \exists z\in \Bbb Z^+\;( z^2=3y^2-y=y(3y-1))\implies$$ $$ \implies\exists a,b \in \Bbb Z^+\;( y=a^2 \land 3y-1=b^2)\implies$$ $$\implies\exists a,b\in \Bbb Z^+\;(3a^2-1=b^2)\implies$$ $$ \implies \exists b\in \Bbb Z^+\;(b^2\equiv -1 \mod 3).$$ The 1st (displayed) line uses the Quadratic Formula.
In the 2nd line, $z\ne 0$ because $y\in \Bbb Z^+\implies y(3y-1)>0.$
In the 3rd line, $a$ and $b$ exist  because $y$ and $3y-1$ are co-prime members of $\Bbb Z^+$ and their product is the square of the positive integer $z.$
