Prove that if $A$ is Hermitian and $A^m=I$, then $A^2=I$ (and $A=I$ if $m$ is odd)

Your approach is right but in the middle it is not clear what you are doing. First since $A$ is hermitian all the eigenvalues of $A$ are real. Further, $A^m = I$ implies that any eigenvalue $\lambda$ satisfies the relation $\lambda^m - 1 =0$. Since $\lambda \in \mathbb{R}$, if $m$ is even, we get that $\lambda = \pm 1$. If $m$ is odd, we get that $\lambda = 1$.

Also, any hermitian matrix permits a decomposition of the form $$A = U \Lambda U^{\dagger}$$ where $U$ is a unitary matrix and $\Lambda$ is a diagonal matrix with the eigenvalues of $A$ as its diagonal entries. This is so since it is possible to find an orthonormal basis of $\mathbb{C}^n$ consisting of $n$ eigenvectors of $A$.

If $m$ is odd, we showed that all the eigenvalues have to be $1$. Hence, $\Lambda$, the diagonal matrix with the eigenvalues of $A$ as its diagonal entries, is the identity matrix i.e. $\Lambda = I$. This gives us that $$A = U I U^{\dagger} = I$$

If $m$ is even, we showed that all the eigen values have to be $\pm 1$. Hence, $\Lambda^2 = I$. This gives us that $$A^2 = (U \Lambda U^{\dagger})(U \Lambda U^{\dagger}) = U \Lambda (U^{\dagger} U) \Lambda U^{\dagger} = U \Lambda I \Lambda U^{\dagger} = U \Lambda^2 U^{\dagger} = UIU^{\dagger} = U U^{\dagger} = I$$

As $\,A\,$ is diagonalizable the only possible forms for its minimal polynomial are $\,(1)\,$ , if $\,z=0\,$, $\,(2)\,$ , if $\,k=0\,$, or $\,(3)\,$ , if $\, k,z\geq 1\,$ . Case $\,(4)\,$is impossible as a square matrix over some field is diagonalizable iff its minimal pol. is the product of different linear polynomials over that field, and from these possibilities either $$A=I\,\,,\,\,A=-I\,\,,\,\,A^2-I=0$$ from which you get what you wanted.