Let's use the notation $(a_1, \ldots, a_m, n)$ to represent a solution. Then we have:

m=4:
(1, 8, 66, 101, 2),
(5, 5, 18, 42, 2),
(5, 11, 59, 200, 3)

m=5:
(1, 1, 45, 133, 261, 2)
(5, 21, 35, 35, 149, 2),
(22, 28, 35, 35, 188, 2)

m=6:
(1, 1, 1, 12, 25, 60, 2),
(3, 3, 3, 3, 58, 98, 2),
(3, 3, 3, 14, 17, 62, 2),
(5, 5, 18, 25, 25, 122, 2),
(2, 2, 13, 13, 21, 138, 3)

m=7:
(1, 1, 1, 1, 5, 17, 39, 2),
(1, 1, 1, 1, 11, 11, 40, 2),
(1, 1, 1, 1, 26, 32, 94, 2),
(2, 2, 6, 6, 6, 13, 55, 2)

Perhaps for $m \geq 4$ there is always a solution for $n=2$ containing $(m-3)$ ones? If so, then this would form the basis of a proof that for $n=2$ there are always solutions for all $m \geq 4$.

**Update** I have used the above construction with $(m-3)$ ones to find new solutions. Here I will use the following format to describe a solution $(m,a_1,a_2,a_3,n)$. Here is what I found:
(8, 1, 12, 27, 2),
(9, 6, 14, 40, 2),
(10, 1, 6, 22, 2),
(11, 1, 1, 16, 2),
(12, 1, 12, 34, 2),
(13, 1, 85, 129, 2),
(15, 1, 575, 687, 2),
(16, 1, 55, 184, 3),
(17, 117, 1273, 1845, 2),
(18, 11, 44, 106, 2),
(22, 1, 34, 82, 2),
(24, 1, 8, 48, 2),
(26, 1, 15, 106, 3),
(27, 1, 30, 265, 5),
(27, 1, 107, 187, 2),
(30, 105, 222, 531, 2).

**UPDATE 2** Instead of ones we can use $(m-3)$ values of $r$. This allows us to find all the missing solutions up to $m=30$. Here I will use the format $(r,m,a_1,a_2,a_3,n)$:
(3, 14, 220, 395, 972, 2),
(4, 19, 41, 510, 861, 2),
(5, 20, 176, 936, 1715, 2),
(2, 21, 203, 406, 965, 2),
(2, 23, 87, 213, 512, 2),
(4, 25, 89, 89, 418, 2),
(3, 28, 13, 1232, 1683, 2),
(3, 29, 91, 325, 741, 2).

Also I was able to find a solution with $n=4$: (1, 34, 43, 43, 442, 4).