$$1, 3, 12, 32,...$$
Above is the sequence of the number of solutions, if there are, to the Diophantine equation : $d^{m+1} =a^{m}+ b^{m}+ c^{m}$ for $m =2$, in positive integers where $a, b$ and $c$ are realtively prime.
For example with the notation: $d^{3} =a^{2}+b^{2}+c^{2} --> (a,b,c,d)$
$$ [1, 1, 5, 3]\\ [3, 19, 31, 11] [9, 17, 31, 11] [19, 21, 23, 11]\\ [1, 19, 139, 27] [1, 71, 121, 27] [5, 83, 113, 27] [7, 95, 103, 27] [11, 49, 131, 27] [17, 25, 137, 27] [23, 55, 127, 27] [25, 37, 133, 27] [29, 41, 131, 27] [43, 47, 125, 27] [43, 85, 103, 27] [59, 89, 91, 27]\\ [1, 5, 207, 35] [1, 75, 193, 35] [1, 93, 185, 35] [1, 135, 157, 35] [3, 29, 205, 35] [5, 57, 199, 35] [9, 137, 155, 35] [11, 27, 205, 35] [15, 103, 179, 35] [15, 143, 149, 35] [19, 67, 195, 35] [25, 43, 201, 35] [25, 111, 173, 35] [27, 89, 185, 35] [33, 95, 181, 35] [33, 115, 169, 35] [37, 59, 195, 35] [43, 135, 151, 35] [47, 129, 155, 35] [51, 55, 193, 35] [51, 125, 157, 35] [55, 103, 171, 35] [61, 123, 155, 35] [65, 97, 171, 35] [65, 137, 141, 35] [67, 75, 181, 35] [71, 103, 165, 35] [73, 135, 139, 35] [79, 97, 165, 35] [79, 115, 153, 35] [103, 125, 129, 35] [109, 113, 135, 35] $$
So, we have 32 different forms of writing $42875$ as the sum of three squares; but only 1 form of expressing $27$ this way, 3 ways to express $11^3$ and 11 to write $3^9$. The sequence seems clearly to be infinite.
And here is the sequence for $m = 3$. It probably still is an infinte sequence :
$$1, 1, 1, 1, 1,..$$
But the sequences for $m \ge 4$ might well be nonexistent. For $m=4$ there might be, perhaps, a seldom very few ones.
The question is, as there always must be a question; find a "1" to $m =4$ and to $m =5$; that is.
P.S : Computing was made using the efficient Pari gp.
The sequence of the number of solutions seems to be : $1, 3, 12, 32, 32, 65, 64, 113, 62, 134,... $ for respective values of d: $3, 11, 27, 35, 51, 59, 75, 83, 99, 107,... $