jmc

Let $d$ be the degree of $P(x).$ Then the degree of $P(P(x))$ is $d^2,$ and the degree of $(x^2+x+1)P(x)$ is $d+2,$ so $$d^2=d+2.$$Then $d^2-d-2=(d-2)(d+1)=0.$ Since $d$ is positive, $d=2.$

Let $P(x)=a{x}^2+bx+c.$ Then $$\begin{array}{rl}P(P(x))& =a(a{x}^2+bx+c{)}^2+b(a{x}^2+bx+c)+c\\ & =a^3{x}^4+2a^{2}b{x}^3+(a{b}^2+2a^{2}c+ab)x^2+(2abc+b^2)x+a{c}^2+bc+c\end{array}$$and $$(x^2+x+1)(a{x}^2+bx+c)=a{x}^4+(a+b)x^3+(a+b+c)x^2+(b+c)x+c.$$Comparing coefficients, we get $$\begin{array}{rl}{a}^3& =a,\\ 2a^{2}b& =a+b,\\ a{b}^2+2a^{2}c+ab& =a+b+c,\\ 2abc+b^2& =b+c,\\ a{c}^2+bc+c& =c.\end{array}$$From $a^3=a,$ $a^3-a=a(a-1)(a+1)=0,$ so $a$ is 0, 1, or $-1.$ But $a$ is the leading coefficient, so $a$ cannot be 0, which means $a$ is 1 or $-1.$

If $a=1,$ then $2b=1+b,$ so $b=1.$ Then $$1+2c+1=1+1+c,$$so $c=0.$ Note that $(a,b,c)=(1,1,0)$ satisfies all the equations.

If $a=-1,$ then $2b=-1+b,$ so $b=-1.$ Then $$-1+2c+1=-1-1+c,$$so $c=-2.$ But then the equation $a{c}^2+bc+c=c$ is not satisfied.

Hence, $(a,b,c)=(1,1,0),$ and $P(x)=\boxed{x^2+x}.$