If $\sigma _{1}(n)\mid \sigma _{2}(n)$, does $n$ has to be a perfect square?

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Let's say $\sigma _{1}(n)\mid \sigma _{2}(n)$. Can we say, therefore $n$ has to be a perfect square? How to show that?

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no

4 sigma_1: 7 sigma_2: 21
9 sigma_1: 13 sigma_2: 91
16 sigma_1: 31 sigma_2: 341
20 sigma_1: 42 sigma_2: 546
25 sigma_1: 31 sigma_2: 651
36 sigma_1: 91 sigma_2: 1911
49 sigma_1: 57 sigma_2: 2451
50 sigma_1: 93 sigma_2: 3255
64 sigma_1: 127 sigma_2: 5461
81 sigma_1: 121 sigma_2: 7381
100 sigma_1: 217 sigma_2: 13671

Here are the nonsquares up to 2000 that work. They do tend to have large square factors.

20 sigma_1: 42 sigma_2: 546
50 sigma_1: 93 sigma_2: 3255
117 sigma_1: 182 sigma_2: 15470
180 sigma_1: 546 sigma_2: 49686
200 sigma_1: 465 sigma_2: 55335
242 sigma_1: 399 sigma_2: 73815
325 sigma_1: 434 sigma_2: 110670
450 sigma_1: 1209 sigma_2: 296205
468 sigma_1: 1274 sigma_2: 324870
500 sigma_1: 1092 sigma_2: 341796
578 sigma_1: 921 sigma_2: 419055
605 sigma_1: 798 sigma_2: 383838
650 sigma_1: 1302 sigma_2: 553350
800 sigma_1: 1953 sigma_2: 888615
968 sigma_1: 1995 sigma_2: 1254855
980 sigma_1: 2394 sigma_2: 1338246
1025 sigma_1: 1302 sigma_2: 1094982
1058 sigma_1: 1659 sigma_2: 1401855
1280 sigma_1: 3066 sigma_2: 2271906
1300 sigma_1: 3038 sigma_2: 2324070
1445 sigma_1: 1842 sigma_2: 2179086
1476 sigma_1: 3822 sigma_2: 3214302
1620 sigma_1: 5082 sigma_2: 4030026
1682 sigma_1: 2613 sigma_2: 3540615
1700 sigma_1: 3906 sigma_2: 3964590
1800 sigma_1: 6045 sigma_2: 5035485
1872 sigma_1: 5642 sigma_2: 5275270