Solving an equation for positive integral values of variables. Question  

Solve
  $$y^3+3y^2+3y=x^3+5x^2-19x+20$$
  for positive integers $x$ and $y$.  

My approach:  

I factorised the equation as
  $(y-x)(x^2+y^2+3x+3y+3)=2[(x-1)(x-10)]$
  and got two ordered pairs of $(x,y)$ i.e. $(1,1);(10,10)$  

I can't go further.  
Thanks for your help.  
PS: Please consider this question without my approach also, because it may mislead you.

Write $$(y+1)^3= y^3+3y^2+3y+1=x^3+5x^2-19x+21$$ 
now, since a discriminant of $5x^2-19x+21$ is $19^2-20\cdot 21<0$ we have
$$x^3<x^3+5x^2-19x+21$$
Also
$$x^3+5x^2-19x+21<x^3+6x^3+12x+8 =(x+2)^3$$
since $x^2+31x-13>0$ for positive $x$. So $$x^3<(y+1)^3<(x+2)^3$$
which is only possible iff $y+1=x+1$ so $x=y$...
