TITLE: How find this $x,y$ such this equation $x^x+x=y!$
QUESTION [15 upvotes]: Find all pairs of positive integers $(x,y)$,such
  $$x^x+x=y!$$

I find the $$(x,y)=(1,2)$$ is such it. and 
$$(x,y)=(2,3)$$
I think this equation have other roots. and maybe use inequality to solve this?
since Use  Bernoulli inequality we  have 
$$x^x=(x-1+1)^x\ge 1+x(x-1)=x^2-x+1$$
Thank you for you help, this problem is from Mathematical contest in china jiangxi province in 2014

REPLY [17 votes]: It's easy to prove that if $x>3$ then $x<y,$ thus $x-1\mid y!,$ but $x^x+x\equiv 1^x+1\equiv 2\pmod{x-1},$ a contradiction.