Solution: Note that \begin{align*} & x^{49} + x^{50} + x^{51} + x^{52} + x^{53} \\ = \ \ & x^{49} \left( 1+x+x^2 \right) + x^{52} \left( 1+x \right) \\ = \ \ & x^{49} \times 0 + x^{52} \left( -x^{2} \right) \\ = \ \ & -x^{54} . \end{align*}

We have \begin{align*} 1+x+x^{2} = 0 & \implies (x-1) \left( 1+x+x^{2} \right) = 0 \\ & \implies x^3 - 1 = 0 \\ & \implies x^3 = 1. \end{align*} Substitute the above, we obtain \begin{align*} x^{49} + x^{50} + x^{51} + x^{52} + x^{53} & = -x^{54} \\ & = -\left( x^3 \right)^{18} \\ & = - 1^{18} \\ & = -1. \end{align*} Therefore, \[ \textcolor{blue}{\boxed{x^{49} + x^{50} + x^{51} + x^{52} + x^{53} = -1}} \]