Not an answer, but maybe a start:
It is fairly clear why trivial cases like $n=18,$ power$=2$ don't work, after all of the sum-pairs $\neq$ a power of $2$ that are $\leq2n$ are stripped away:
Complete cycles are much easier to search for: cycleP[33, 2] (for $n=33,$ power$=2$, code below) produces
whereas cyclePall[23, 2] produces
and it is clear why nothing below $300$ish will work for power $3$ by just looking at dangling nodes of $n=200,$ power$=3$:
cycleP[n_, pow_] :=
With[{graph = Graph[DeleteDuplicates[Flatten[Thread[#[[1]] -> #[[2]]] & /@
Transpose[{Range@n, Table[If[#[[1]] == hh, #[[2]], #[[1]]] & /@
Select[Flatten[DeleteCases[Table[With[{aa = Transpose@{(ConstantArray[#, #]
&@nn - Range@nn), Reverse@(ConstantArray[#, #] &@nn - Range@nn)}},
Select[Rest@ Take[aa, Floor[Length@aa/2]], #[[1]] <= n && #[[2]] <= n &]],
{nn, Range[2, Floor[(2 n)^(1/pow)]]^pow + 1}], {}], 1], #[[1]] == hh \[Or] #[[2]]
== hh &], {hh, n}]}]], Sort[#1] == Sort[#2] &], DirectedEdges -> False,
VertexLabels -> "Name"]}, Column[{Show[#, ImageSize -> 400] &@
HighlightGraph[graph, Style[FindCycle[graph, {n}], {Darker@Red, Thick}]],
Flatten@(#[[All, 1]] & /@ FindCycle[graph, {n}])}]]
cyclePall[n_, pow_] :=
With[{cc = DeleteDuplicates[Flatten[Thread[#[[1]] -> #[[2]]] & /@
Transpose[{Range@n, Table[If[#[[1]] == hh, #[[2]], #[[1]]] & /@
Select[Flatten[DeleteCases[Table[With[{aa = Transpose@{(ConstantArray[#, #] &@nn -
Range@nn), Reverse@(ConstantArray[#, #] &@nn - Range@nn)}},
Select[Rest@ Take[aa, Floor[Length@aa/2]], #[[1]] <= n && #[[2]] <=
n &]], {nn, Range[2, Floor[(2 n)^(1/pow)]]^pow + 1}], {}], 1], #[[1]] == hh \[Or]
#[[2]] == hh &], {hh, n}]}]], Sort[#1] == Sort[#2] &]}, With[{dd =
Split@Sort@Join[cc[[All, 1]], cc[[All, 2]]]},
With[{jj = DeleteCases[Flatten@(If[Length@# == First@Sort[Length@# & /@ dd], #, 0]
& /@ dd), 0]}, With[{ll = Flatten@Table[Thread[#[[1]] -> #[[2]]] & /@
Transpose@{ConstantArray[jj[[kk]], n], Range@n}, {kk, Length@jj}]},
With[{zz = Table[Join[{ll[[vv]]}, cc], {vv, Length@ll}]}, With[{zzz =
DeleteCases[Table[FindCycle[Graph[zz[[ww]], DirectedEdges -> False,
VertexLabels -> "Name"], {n}], {ww, Length@zz}], {}]}, With[{graphs =
(HighlightGraph[Graph[cc, DirectedEdges -> False, VertexLabels -> "Name"],
Style[#, {Darker@Red, Thick}]] & /@ zzz)},Column[{If[Length@graphs == 0,
Show[Graph[cc, DirectedEdges -> False, VertexLabels -> "Name"], ImageSize -> 400],
Show[#, ImageSize -> 400] & /@ graphs],#[[All, 1]] & /@ (Rest@# & /@
Flatten[zzz, 1])}]]]]]]]]
(Mathematica 10 only)
#Update#
This is a placeholder, as I have no access to my computer for a few days, but outlines rough sketch of idea:
Illustrated as a clock with $1$ at the top for $n=24$ power$=2,$ this is a far more systematic way to think about it:
grouping sum-pairs by square $\leq 48.$ The superimposition shows all possible paths with one remaining odd "leg". There is therefore no Eulerian circuit (or trail) for $n=24$ power$=2.$
As long as there is at least two adjoining points for all but $2$ points, there will be an Eulerian trail. If all points have at least two adjoining points, it is likely there will be an Eulerian circuit.
It is highly likely then (and not too difficult to prove, I should have thought), that over a certain $n$ for each power will guarantee at least one Eulerian trail.