Consider $\mathbb{C}$-case. We will need two following simple observations.
1) Let $a=a_1\cdot\ldots\cdot a_n$, for some elements $a$,$a_1,\ldots,a_n$ of algebra $A$ and assume that $a_1,\ldots,a_n$ commute. Then $ a\text{ is invertible }\Longleftrightarrow a_1,\ldots, a_n\text{ are invertible.} $
2) One can easily check that $ T^n-\lambda^nI=(T-\lambda I)\left(\sum\limits_{k=0}^{n-1}\lambda^{n-1-k}T^k\right)=\left(\sum\limits_{k=0}^{n-1}\lambda^{n-1-k}T^k\right)(T-\lambda I) $
Now we have implication $ \lambda^n\notin\sigma_\mathbb{C}(T^n)\Longleftrightarrow T^n-\lambda^nI\text{ is invertible }\Longrightarrow $ $ T-\lambda I\text{ is invertible }\Longleftrightarrow \lambda\notin\sigma_\mathbb{C}(T) $
Thus, $\lambda\in\sigma_\mathbb{C}(T)\Longrightarrow\lambda^n\in\sigma_\mathbb{C}(T^n)$. As Robert Israel pointed out the reverse implication holds only if we are given $\lambda^n\in\sigma_\mathbb{C}(T^n)$ for all $n\in\mathbb{N}$.
Consider $\mathbb{R}$-case. Define bounded linear operators $ T:\mathbb{R}^2\to\mathbb{R}^2:(x_1,x_2)\mapsto(-x_2,x_1) $ $ T^2:\mathbb{R}^2\to\mathbb{R}^2:(x_1,x_2)\mapsto(-x_1,-x_2) $ This are rotations on the angles $90^\circ$ and $180^\circ$ respectively. Matrices of this operators in standard basis are $ [T]=\begin{pmatrix}0&&-1\\1&&0\end{pmatrix}\qquad [T^2]=\begin{pmatrix}-1&&0\\0&&-1\end{pmatrix} $ You can easily check that $\sigma_\mathbb{R}(T)=\varnothing$ whereas $\sigma_\mathbb{R}(T^2)=\{-1\}$.