Divisor of $x^2+x+1$ can be square number? $$1^2+1+1=3$$
$$2^2+2+1=7$$
$$8^2+8+1=73$$
$$10^2+10+1=111=3\cdot37$$
There is no divisor which is square number.
Is it just coincidence? Or can be proved?
*I'm not english user, so my grammer might be wrong

What about $x=653$, where 
$$x^2+x+1=427063=7\cdot (13\cdot 19)^2\,?$$
How did I find this $x$?  I first suppose that $x^2+x+1=ay^2$ for some positive integers $a$ and $y$.  This means
$$(2x+1)^2-a(2y)^2=-3\,.$$
Let $u:=2x+1$ and $v:=2y$.  Then, we are to solve the Pell-type equation $$u^2-av^2=-3\,,$$ where $u,v\in\mathbb{Z}_{>0}$.  In particular, the case $x=2$ yields $a=7$ and $y=1$.  Therefore, I attempt to solve
$$u^2-7v^2=-3\,,\tag{*}$$
where a minimal solution is $(u,v)=(5,2)$.  Since $u^2-7v^2=1$ has the minimal solution $(u,v)=(8,3)$, we obtain an infinite family of solutions $(u,v)$ of (*):
$$u+v\sqrt{7}=(5+2\sqrt{7})\,(8+3\sqrt{7})^k\,,\tag{#}$$
where $k$ is a positive integer.  We want $u$ to be odd, so $k=1$ does not work.  With $k=2$, we get $(u,v)=(1307,494)$, so $$x=\frac{1307-1}{2}=653\text{ and }y=\frac{494}{2}=13\cdot 19$$
form a counterexample.  (Using $k=-2$, we get lhf's counterexample $x=18$.  Indeed, with even values of $k$ in $(\#)$, we obtain infinitely many counterexamples.)
