Let $A \in\mathrm{SL}(2,\mathbb{C})$ and $p > 1$ be a natural number. Under which conditions the following statement is true?
$A^p$ is a diagonal matrix implies $A$ is a diagonal matrix
For the case $p=2$ and $A^2 \neq \pm \mathrm{Id}$ it is a simple calculation to show that it is true.
Even when $p=2$ and $A^2\neq \pm I_2$, this is false. So I'm not sure what your computations were. Take $$ A=\pmatrix{0&\lambda\\1&0}\qquad \qquad A^2=\pmatrix{\lambda&0\\0&\lambda}. $$ As soon as $A^p$ has a repeated eigenvalue, we can build a counterexample. In other words, over $\mathbb{R}$ or $\mathbb{C}$, $\lambda I_m$ has a nondiagonal $p$th root for every $n\geq 2$ and $p\geq 2$. As mentioned by Georges Elencwajg and alex.jordan, it suffices to scale the appropriate rotation ($2\pi/p$, unless $p$ is even, $\lambda<0$, and you really want to do this in $\mathbb{R}$, in which case you should scale the rotation of angle $\pi/p$).
But if $A^p$ has no repeated eigenvalue, in $M_n(K)$ and for any $p\geq 2$, we have a sufficient condition.
Proof: since $A$ commutes with $A^p$, it leaves the eigenspaces of $A^p$ invariant. By assumption, these are one-dimensional. So they must be eigenspaces for $A$ as well. QED.
More generally, if for some polynomial $p(X)\in K[X]$, $p(A)$ is diagonalizable with pairwise distinct coefficients, then $A$ is diagonalizable in the same basis.
Of course, all this falls apart as soon as $p(A)$ has an eigenvalue of multiplicity at least two.
Remark: it is good to think about this in terms of commutant. Given a diagonalizable matrix $B\in M_n(K)$, denote $C(B):=\{A\in M_n(K)\,;\,AB=BA\}$. This is a unital subalgebra of $M_n(K)$. When $B$ is in diagonalized form, $C(B)$ is block diagonal with a block $M_{n_j}(K)$ corresponding to each eigenvalue $\lambda_j$ of multiplicity $n_j$. So when all the eigenvalues of $B$ have multiplicity $1$, all these blocks reduce to scalars, and the commutant is the algebra all diagonal matrices in this basis. In your case, $A$ belongs to the commutant of $B=A^p$.