Yes, very similar to Cone Sul's one..! to solve, see the perfect square a² more closer to n, and do the substituition n = a²+r, with r < 2a+1.

$n=x^{2}+3x+3,x=1,2,4.$

Yes. Let $a=[\sqrt n ]$. Then $n=a^{2}+r,0\le r\le 2a.$ From $a+1|n-1=(a+1)(a-1)+r$ we get $a+1|r$, because $r\le 2a$ $r=a+1$ or $r=0$ (I forgot about r=0). From $a-1|n+1$ we get $a-1|r+2$. If $r=0$ $n=4$ or $n=9$. If $r=a+1$ we get $x=a-1|4, n=x^{2}+3x+3, x=1,2,4.$