On the equation involving the sum of divisors function $\sigma(105n+\sigma(n))=108\sigma(n)$ I am curious about the solutions of the following equation involving the sum of divisors function $\sigma(m)=\sum_{d\mid m}d$
$$\sigma(105n+\sigma(n))=108\sigma(n).\tag{1}$$
It is obvious, since $107$ isn't a Mersenne prime, that every even perfect number is a solution of our equation $(1)$. And if we presume that there exist some odd perfect numbers coprime with the prime $107$, these should be also solutions of our equation. 
Computational fact. Our sequence starts as $$6,28,402,496,1512,1710,1876,7980,8128,15012,29967,30267,\ldots$$
that you can see using Sage Cell Server (choose GP as language) with this code 
for (i = 1, 1000000,if(sigma(105*i+sigma(i))==108*sigma(i),print(i)))

Question. Is it possible to prove that the equation
  $$\sigma(105n+\sigma(n))=108\sigma(n)$$
   has infinitely many solutions? What work can be done*? Many thanks.

*Since I think that this is a difficult question (this kind of equations involving arithmetic functions are difficult, I should to accept an answer showing what work can be done, in the case that a full answer isn't feasible). Thus I am asking if you can to show an infinitude of solutions, or a compelling argument about why we can build such sequence.
 A: Let 
$$\mu(n) = \sigma(n)+105n,\tag1$$
$$\nu(m) = \dfrac1{105}\left(m-\dfrac{\sigma(m)}{108}\right),\tag2$$
then the issue equality can be presented in the form of
$$\nu(\mu(n)) = n.\tag3$$
In particular, for perfect $n$ in the case $\gcd(n, 107) = 1$
$$\mu(n) = \sigma(n) + 105n = 107n,$$
$$\nu(\mu(n)) = \dfrac1{105}\left(107n - \dfrac{\sigma(107n)}{108}\right) = \dfrac1{105}\left(107n - \dfrac{(107+1)2n}{108}\right) = n.$$
Formula $(2)$ allows to look for possible values of $m=\mu(n)$ using the system 
$$108\ |\ \sigma(m),\quad 105\ |\ m - \dfrac {\sigma(m)}{108},\tag4 $$
and then verify them by the solving of the equation
$$\sigma(n)+105n = m.$$
One can see that the possible solutions of the system $(4)$ can be presented in the forms of 
$$m = 108k-1 \in \mathbb P,\quad m = 107, 431, 641, 863, 971\dots\tag5$$
$$m = 85k,\quad \gcd(k, 85) = 1,\quad m = 85, 170, 255, 340, 510, 595\dots \tag 6$$
$$m = 102k,\quad \gcd(k, 102) = 1,\quad m=102, 510, 714, 1122\dots \tag 7$$
and some others ($m= 23k,47k, 71k$), according to the factorization of $108$ in sigma-functions.
This way can be helpful as some start ideas for analytic investigation of this hard topic.
