smc

Let $z=s$ be a root of $z^2+z+1$ so that $s^2+s+1=0.$ It follows that $$(s-1)\left(s^2+s+1\right)=s^3-1=0,$$ from which $s^3=1,$ but $s\ne 1.$ Note that $$\begin{array}{rl}{s}^{2021}+1& =s^{3\cdot 673+2}+1\\ & =(s^3{)}^{673}\cdot s^2+1\\ & =s^2+1\\ & =\left(s^2+s+1\right)-s\\ & =-s.\end{array}$$ Since $z^{2021}+1=-z$ for each root $z=s$ of $z^2+z+1,$ the remainder when $z^{2021}+1$ is divided by $z^2+z+1$ is $R(z)=\boxed{-z}.$