Let $\sigma(x)$ be the sum of the divisors of $x$.
If $\sigma(N) = aN + b$, where $\gcd(a, b) = 1$, $a \geq 2$, and $b$ could be negative, does it follow that at least one of $N$'s factors is solitary?
Of course, we exclude the case $p \mid N$ or $q^k \mid N$, where $p$ and $q$ are primes, and $k \geq 1$, as it is already known that primes and powers of primes are solitary.
(This is only a partial answer to the question.)
If $\sigma(N) = {a_1}N + 1$ for some $a_1 \geq 2$, then $\gcd(a,b)=\gcd(a_1,1)=1$. Additionally, $\gcd(N,\sigma(N))=\gcd(N,{a_1}N+1)=1$, since $\sigma(N)-{a_1}N=1$ expresses $1$ as a linear combination of $\sigma(N)$ and $N$. Therefore, $N$ itself is solitary.
A similar result holds for $M$ when $$\sigma(M) = {a_2}M - 1.$$
The situation when $b = 0$ leads to the study of (multi)perfects, and (therefore?) remains open. Similarly, the case $|b| > 1$ appears (?) to be intractable.
(I'm flagging this as community wiki in case somebody else is able to settle the remaining cases.)