If $z^2 + z + 1 = 0,$ find
\[z^{49} + z^{50} + z^{51} + z^{52} + z^{53}.\]
Since $z^2 + z + 1 = 0,$ $(z - 1)(z^2 + z + 1) = 0,$ which simplifies to $z^3 = 1.$  Therefore,
\[z^{49} = (z^3)^{16} \cdot z = z.\]Then
\begin{align*}
z^{49} + z^{50} + z^{51} + z^{52} + z^{53} &= z + z^2 + z^3 + z^4 + z^5 \\
&= z + z^2 + 1 + z + z^2 \\
&= z + z^2 = \boxed{-1}.
\end{align*}