<s>[INST] Perfect numbers $n$ such that $2^k(n+1)$ is also perfect The smallest two perfect numbers $n=6$ and $m=28$ satisfy
$$
\frac{m}{n+1} = 2^k
$$
with $k=2.$
Question: Are there more pairs of perfect numbers $n,m$  with $n < m$
and such that
$$
\frac{m}{n+1} = 2^k
$$
for some positive integer $k>0.$
Observe that the perfect number $n$ , the smallest of $n,m$
may be also an odd number.
 [/INST] If $m$ is odd, it's clearly impossible. 
If $m$ is even and $n$ is odd, I don't know. 
So suppose $m$, $n$ both even. Then $m=2^{r-1}p$ where $p=2^r-1$ is prime, and $n=2^{s-1}q$ where $q=2^s-1$ is prime, and $s\lt r$. 
The equation becomes $$2^k(n+1)=2^k(2^{s-1}q+1)=2^{k+s-1}q+2^k=2^{r-1}p$$ Now $2^k$ divides the second last term, so it divides the last term, so $2^{s-1}q+1=2^{r-k-1}p$. If $s\gt1$ this forces $r-k-1=0$, so $2^{s-1}q+1=p=2^r-1$. Then $2^r-2^{s-1}q=2$, so either $r\le1$ or $s\le2$. But $s\lt r$, so we reject $r\le1$, so $s=2$, $q=3$, and there's only the one solution. 
</s>