A beautiful problem! Take m_1<m_2<... such that all prime factors of a^m_k-1 are in A for any k. Lemma 1: The set of prime factors of (m_k)k>=1 is finite. Proof: Suppose we can find p_1<p_2<... prime factors of the elements m_1,m_2,.... Then for any k the prime factors of a^p_k-1 are in A. Since A has a finite number of subsets, it follows that we can find an infinite set B such that whenever i and j are in B, a^p_i-1 and a^p_j-1 have the same prime factors. Since if i<>J we have (p_i,p_j)=1 ( and thus (a^p_i-1,a^p_j-1)=1), it follows that any prime factor of a^p_i-1 (with i in B) is a divisor of a-1. Take i in B such that p_i>a-1 and take q a prime dividing 1+a+...+a^(p_i-1). Then q|a^p_i-1 and thus q|a-1. Thus, q|p_i and q=p_i>a-1, contradiction with a previous statement. Thus, we can find primes p_1,...,p_k (not those from the lemma!) such that each m_i has its prime factors between them. If we denote by x(n,i) the exponent of p_i in m_n, then there is i such that (x(n,i))n>=1 is unbounded. Thus, there is a prime p and an unbounded sequence b_n such that the set of prime factors of a^p^b_n-1 is finite. Take a number s>1 and suppose q is a prime factor of 1+a^p^(s-1)+...+a^p^(s-1)*(p-1). Then q|a^p^s-1 and thus the order of a modulo q is p^r, with r<=s. If r<s, then it easily follows that q|p and thus q=p. Otherwise, p^s|q-1 and thus q>p^s. Since the sequence b_n is unbounded and the set of prime factors of a^p^b_n-1 is bounded, there is n_o such that for all n>n_0 1+a^p^(b_n-1)+...+a^p^(b_n-1)(p-1) is a power of p. But I really don't think this is possible,.

A small missprint: (a^p_i-1,a^p_j-1)=a-1. And the solution to the last statement: Suppose we have 1+x+...+x^(p-1)=p^t, with t>1. Then (x-1)p^t=x^p-1. From Fermat theorem we have p|x^p-x, thus p|x-1. Write x=1+rp^u, with (r,p)=1. Then rp^(u+t)=rp^(u+1)+Multiple of p^2u. Impossible (it follows that r is divisible ny p!). And the last statement is proved.

I have another solution: We will use two lemmas: Lemma 1:If a>1 and q is a prime divisior of a ^p -1 which p is a prime,then p/q-1. Lemma 2: Every prime divisior of (a ^{p ^m+1} -1)/(a ^{p ^m}-1) have this form:p ^m+1 *k+1.(m,k is a positive integer.) Use little's Fermat theorem,we can easily prove two lemmas. B is the set of all positive integer m which satisfied:every prime divisior of a ^m -1 belongs to A. with this problem,I suppose for contradiction,card B is infinite.We note that :if m belongs to B then every divisior of m belongs to B.So we really needto all the prime and it's power in B.If in B ,there have finite prime in B:then exits infinite sequences:p,p ² ,..,p ⁿ ,...(which p is a prime belongs B) are belongs to B.By lemma 2,we get contradiction because the number of all divisior of (a ^{p ^m+1} -1)/(a ^{p ^m}-1) is infinite.So,There have infinite prime in B.Note that: every prime q' in B there exits a prime p' in A which p'/a ^q' -1 so by lemma 1,q'/p'-1 so card A is infinite,contradiction. That's all.if I mistake,please sorry to everybody .Thanks!