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

The square of any prime $p\equiv1\pmod3$ appears as a factor of $x^2+x+1$ for some choice of $x$.
This is seen as follows. 
The multiplicative group $\Bbb{Z}_{p^2}^*$ of coprime residue classes modulo $p^2$ is known to be cyclic of order $p(p-1)$. It follows that there is an element of order three in that group. Let the residue class of $x$ be one such. Because $p>3$ the order of $x$ modulo $p$ is also equal to three. Implying that $x-1$ is not a multiple of $p$. But $$x^3-1=(x-1)(x^2+x+1)\equiv1-1=0\pmod{p^2}$$
by construction,
so we can conclude that $$x^2+x+1\equiv0\pmod{p^2}.$$

A non-deterministic way of finding such an $x$ is to take a random integer $a$, and calculate the remainder $x$ of $a^{p(p-1)/3}$ modulo $p^2$. If the result is $x\neq1$, then we have found the required element of order three.
For example, with $p=31$, $a=3$ we find that 
$$
3^{31(31-1)/3}=3^{310}\equiv521\pmod{31^2}.
$$
And with $x=521$ we get
$$
521^2+521+1=271963=31^2\cdot283
$$
as promised.

On the other hand no prime $p\equiv-1\pmod3$ will appear as a factor of $x^2+x+1$ for any integer $x>1$. This is because the factorization $(x^3-1)=(x-1)(x^2+x+1)\equiv0\pmod p$ implies that $x$ has order $1$ or $3$ in the group $\Bbb{Z}_p^*$. In the former case $x\equiv1\pmod3$ and therefore 
$x^2+x+1\equiv1+1+1=3\not\equiv0\pmod p$. In the latter case Lagrange's theorem from elementary group theory tells us that $3$ must be a factor of the order of the group $G=\Bbb{Z}_p^*$. As $|G|=p-1$ we can conclude that $p\equiv1\pmod3$.

By more or less the same argument we can show that for all $p\equiv1\pmod3$ the number $x^2+x+1$ can be made divisible by any power $p^k$. This time the group has order $p^{k-1}(p-1)$. As an example consider $p=7,k=5$. The above method produces $$3^{7^4(7-1)/3}\equiv1353\pmod{7^5}.$$ And, predictably,
$$1353^2+1353+1=7^5\cdot109.$$
