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{align*} 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{align*}and so on. Thus, the powers of $ z $ reduce to 1, $ z, $ and $ z^2, $ in cycles. Also, \begin{align*} \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{align*}Since the powers of $ z $ reduce to 1, $ z, $ and $ z^2, $ in cycles, \begin{align*} \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 + \dots + \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{align*}