Infinitely many solutions to an equation with primes: $pqr+22=s$? (open problem) The question that I am talking about is:

A natural number $n$ has property $(*)$, if $n=pqrs$, product of $4$ distinct primes, and, if we order its $16$ divisors, i.e. $$1=d_1<d_2<...<d_{16}=n,$$we have: $d_9-d_8=22$.
Are there infinitely many numbers $n$, that satisfy $(*)$ ?

In fact, by discussing the cases of $s$, compared with the other primes, the above problem is equivalent to finding solutions to one of the following equations: ($p<q<r<s$)
$1) s=pqr+22$, if $pqr<s$;
$2) s=pqr-22$, if $qr<s<pqr$;
$3) ps=22+qr$, if $qr/p<s<qr$;
$4) qr=22+ps$, if $s<qr/p$.
If any of $1)-4)$ has infinitely many solutions, then the original problem is solved.
My attempt to the first one, treated separately:
First of all, it is easy to notice that $p,q,r \neq 2,11.$
I tried to reduce the degrees of freedom, by setting, for example, $p=3,q=5$, in order to get some sort of arithmetic progressions, but I don't know if Dirichlet's theorems can help.
P.S. The above question is derived from $1995$ Irish Math Olympiad problem. At that time, the task was similar to finding the minimum solution, that is $n=3\cdot 5 \cdot 7 \cdot 19$, which is a solution to $3)$. My apologies for asking such a difficult question.
IMPORTANT EDIT: The problem is still open, although I have accepted an answer (or comment), for being the most relevant.
 A: THIS IS NOT AN ANSWER, it is just a comment maybe not impertinent and it is put here for obvious reasons of space.
It is clear that $2$ must be discarded. For all pair of odd primes $p,q$ the diophantine equation $px-qy=22$ has infinitely many integer solutions in such a way that there are infinitely many primes for $x$ and $y$ (Dirichlet). The point is that for our purpose we need these primes appear  simultaneously (which could deserve to be an interesting problem maybe). We look at some examples.
$$3x-5y=22\iff(x,y)=(5n+4,3n-2)\Rightarrow n=3 \space\text {gives }(r,s)=(19,7)\\5x-7y=22\iff(x,y)=(7n+3,5n-1)\Rightarrow n=4 \space\text {gives }(r,s)=(31,19)\\13x-17y=22\iff(x,y)=(17n+3,13n+1)\Rightarrow n=4 \space\text {gives }(r,s)=(19,7)$$
It must be said that it could be the case that only one of this class of equations is sufficient (to show which would ultimately be a refinement of Dirichlet's Theorem). Another thing is that we have not taken into account the required eighth and ninth position but have limited ourselves to finding four odd primes.
A: Comment:
Based on finding of OP by computer using prime generator $15m+22$ and my comment that all primes are of the form $30k+r$ where $r=1, 7, 11, 13, 17. 19, 23, 29$ we solve following Diophantine equation for every r:
$30k+r=15m+22$$\space\space\space\space\space\space\space (1)$
Here for example I solve it for $r=7$:
$p=30k+7=15m+22$
Or:
$2k-m=1$
General form of solutions are:
$(k, m)=(t+2, 2t+3)$
Such that:
$p=30(t+2)+7=15(2t+3)+22=30t+67$
In this way prime generator $p=30t+67$ can give primes satisfying the condition, some examples:
$t=1\rightarrow p=107=22+5\times17$
$t=2\rightarrow p=127=22+5\times 19$
We may try other values for r and solve equation (1) such that the result is $p=22+pqr$ and show experimentally that there can be infinite such primes.
