jmc

Since $z^2+z+1=0,$ $(z-1)(z^2+z+1)=0.$ This expands as $z^3-1=0,$ so $z^3=1.$ Then $$\begin{array}{rl}{z}^4& =z\cdot z^3=z,\\ z^5& =z\cdot z^4=z^2,\\ z^6& =z\cdot z^2=z^3=1,\\ z^7& =z\cdot z^6=z,\\ z^8& =z\cdot z^7=z^2,\\ z^9& =z\cdot z^8=z^3=1,\end{array}$$and so on. Thus, the powers of $z$ reduce to 1, $z,$ and $z^2,$ in cycles.

Also, $$\begin{array}{rl}{\left(z+\frac{1}{z}\right)}^2& =(z+z^2{)}^2=(-1{)}^2=1,\\ {\left(z^2+\frac{1}{z^2}\right)}^2& =(z^2+z{)}^2=(-1{)}^2=1,\\ {\left(z^3+\frac{1}{z^3}\right)}^2& =(1+1{)}^2=4.\end{array}$$Since the powers of $z$ reduce to 1, $z,$ and $z^2,$ in cycles, $$\begin{array}{rl}{\left(z+\frac{1}{z}\right)}^2+{\left(z^2+\frac{1}{z^2}\right)}^2+{\left(z^3+\frac{1}{z^3}\right)}^2+\cdots +{\left(z^{45}+\frac{1}{z^{45}}\right)}^2& =15\left[{\left(z+\frac{1}{z}\right)}^2+{\left(z^2+\frac{1}{z^2}\right)}^2+{\left(z^3+\frac{1}{z^3}\right)}^2\right]\\ & =15(1+1+4)=\boxed{90}.\end{array}$$