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.

[After typing out this attempt at a "partial" answer, I realized that the details have already been worked out by Luis, Gerhard and Todd.  I am posting it as an answer for anybody else who might be interested in how the final result is obtained. - Arnie]
Suppose $m$ is even and $n$ is odd.
Then if $m$ and $n$ are perfect numbers, we have the forms
$$m = 2^{p-1}({2^p} - 1),$$
where $p$ and ${2^p} - 1$ are primes, and
$$n = {q^r}{s^2},$$
where $q$ is prime with $q \equiv r \equiv 1 \pmod 4$ and $\gcd(q,s) = 1$.
Now, from the additional constraint
$$\frac{m}{n + 1} = 2^k,$$
where $k > 0$ is an integer, we obtain the equation
$$m = {2^k}(n + 1).$$
Writing out this last equation in full by plugging in the respective forms for $m$ and $n$ as before, we get
$$2^{p-1}({2^p} - 1) = {2^k}({q^r}{s^2} + 1).$$
By divisibility considerations, since we can assume without loss of generality that $p \geq 2$, $k \geq 1$ and $r \geq 1$, and since $n \equiv 1 \pmod 4$, we get
$$\gcd(2^{p-1}, {q^r}{s^2} + 1) = 2 \Longrightarrow 2^{p-2} \mid 2^k.$$
Similarly, $\gcd({2^k},{2^p} - 1) = 1$ and $({q^r}{s^2} + 1) = n + 1 \equiv 2 \pmod 4 \Longrightarrow 2^{k+1} \mid 2^{p-1} \Longrightarrow 2^k \mid 2^{p-2}$.
Thus,
$$2^k = 2^{p - 2}.$$
This gives
$$k = p - 2.$$
Consequently, we get
$$\frac{m}{n + 1} = 2^k = 2^{p-2} = \frac{2^{p-1}({2^p} - 1)}{{q^r}{s^2} + 1},$$
which yields
$$\frac{1}{2} = \frac{2^p - 1}{{q^r}{s^2} + 1}.$$
This implies
$${q^r}{s^2} + 1 = 2^{p+1} - 2.$$
This finally gives (as Gerhard, Todd and Luis had already noted)
$${q^r}{s^2} = n = 2^{p+1} - 3.$$
