$x=ad,y=bd, (a,b)=1$ $abd^2=a+b+d$ $abd^2=a+b+d \leq 2ab+abd \to d^2 \leq 2+d \to d \leq 2$ $d=1: ab=a+b+1 \to (a-1)(b-1)=2 \to (a,b)=(2,3) \to (x,y)=(2,3)$ $d=2: 4ab=a+b+2 \to (a,b)=1 \to (x,y)=(2,2)$

$x = ad, y = bd, \gcd(a,b) = 1$ We have: WLOG $x \geq y$ $abd^3 = ad + bd + d^2$ $xy = abd^2 = a + b + d \leq 3x$ $3 \geq y$ Now just check cases and done

Rewrite the equation as $d^2 -xyd+(x+y)=0.$ $d=\frac{xy\pm\sqrt{x^2 y^2 -4(x+y)}}{2}.$ Can we proceed from here?

We can use inequality logic to my opinion.Call $x=da$,$y=db$,where the equation becomes $d(dab-1)=a+b$.$d|a+b$ from here,call $a+b=dk$,where $k \leq a+b$ surely.We now have $dab=k+1$ and $dab \leq a+b+1$.However,$ab \geq a+b+1$ except the cases: 1)where one of the $a$ and $b=1$,2)$a=2,b=2$ and 3)$(a,b)=2,3$.We'll analyse cases.First,let's start with 3).We have $d(6d-1)=5$ $\Rightarrow$ 3.1)$d=1$ or 3.2)$d=5$.Surely,3.2) is false,however we get the $(x,y)=2,3$ solution from 3.1). 2) is easy as well,no solution from there. Coming to 1),let's assume that $b=1$.We get $ad^2 \leq a+d+1$.So,again we have $ad\leq a+d$ for special cases.Analysing those special cases gives $d=2$,$(a,b)=1$,so,$(x,y)=2,2$ as solution.