For any positive integer $m>1$, let $\sigma'(m)$ denote the sum of positive divisors of $m$, excluding $m$. So, we have to find all positive integer $n>1$ that $\sigma'(n)=n-1$ and $\sigma' (n^m)=n^m-1$ for some $m\in \mathbb{Z}_{>1}$. We have $\sigma' (n^m) = \sigma' (n) \times \Big( n^{m-1}+n^{m-2}+...+n+1\Big)$. Note that for each positive divisor $d<n$ of $n$, $dn^i$ is a positive divisor of $n^m$ which is strictly less than $n^m$ for all $i\in \{ 0,1,2,...,m-1\}$. Also, there not exist pairs $(d_1,i_1)\neq (d_2,i_2)$ where $d_1,d_2$ are positive divisors of $n$ and $i_1,i_2\in \{ 0,1,2,...,m-1\}$ that $d_1n^{i_1}=d_2n^{i_2}$ (by taking $v_n$ on both sides, note that $d<n$, we get $i_1=i_2$ and so $d_1=d_2$). This gives us $dn^i$ where $d<n$ is a positive divisor of $n$ and $i\in \{ 0,1,2,...,m-1\}$ are pairwise distinct positive divisor of $n^m$ which is strictly less than $n^m$. This gives us $\sigma' (n^m) \geq \sigma' (n) \times \Big( n^{m-1}+n^{m-2}+...+n+1\Big)$. Hence, the equality must holds, which means there's no other divisor of $n^m$ which is streictly less than $n^m$ that is not in that form. If $n$ have at least two prime divisors, let one of them be $p$. We get that $p^{mv_p(n)}\neq n^m$ is a divisor of $n^m$. We have $p^{mv_p(n)}=dn^i$ for some divisor $d<n$ of $n$ and $i\in \{ 0,1,2,...,m-1\}$. For any other prime divisor $q$ of $n$, we have $0=v_q\Big( p^{mv_p(n)}\Big) =v_q(dn^i)=v_q(d)+iv_q(n)\Rightarrow i=0$. Hence $d=p^{mv_p(n)}\mid n\Rightarrow mv_p(n)\leq v_p(n)\Rightarrow v_p(n)=0$, contradiction. So, $n$ must be a prime power, let $n=p^t$ for a prime number $p$ and positive integer $t$. We have $\sigma' (n)=\frac{p^t-1}{p-1} =n-1=p^t-1\Rightarrow p=2$. Easy to see that all power of $2$ satisfy the condition, hence is the answer, done.