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I'm not sure if this question has the appropriate level for MO. If not, feel free to vote for closing.

Let $m \ge 2$. Are there odd primes $p_1 \le ... \le p_m$ and non-negative integers $n_1,...,n_m$ such that $$\prod_{i=1}^m (p_i^{n_i}-1) \quad \text{ divides }\quad (\prod_{i=1}^m p_i^{n_i})-1 \quad\quad ?$$

Background: If there aren't such primes, that would answer a conjecture in another question affirmatively (if one requires $|A|$ to be odd):

A question on finite commutative rings

So far, I could show that for $m=2$ the described constellation isn't possible. By writing $$ \prod_{i=1}^m a_i \quad \text{divides}\quad \prod_{i=1}^m (a_i + 1) -1 \quad ?$$ , I think the left hand side and the right hand side are too close together such that $\prod_{i=1}^m a_i$ can divide the difference. But I may be wrong.

Edit: Thank you all very much for the comments.

My statement about $m=2$ wasn't quite correct, since $(3 -1) \cdot (3-1) \mid 3 \cdot 3 -1$ is a (the only) solution.

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    $\begingroup$ @A Haynes: You meant that $\varphi(n)$ divides $n-1$ (not $n$). $\endgroup$
    – GH from MO
    Commented Nov 24, 2011 at 17:38
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    $\begingroup$ Another solution for $m=4$ is $(3-1)^4 | 3^4-1$. $\endgroup$
    – François Brunault
    Commented Nov 24, 2011 at 19:47
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    $\begingroup$ Another solution for $m=7$ is $3,3,3,3,5,5,89$ (with all $n_i=1$). $\endgroup$
    – François Brunault
    Commented Nov 24, 2011 at 20:10
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    $\begingroup$ Yet another solution (this time with prime powers) : $5, 5, 9, 9, 89$. $\endgroup$
    – François Brunault
    Commented Nov 24, 2011 at 21:15

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