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 $\sigma_\mathbb{R}(T^2)=\{-1\}$.