Get a List With `Partition`

Clear["Global`*"]

$HistoryLength = 0;

The two methods produce identical results

With[{n = 1000},
 List /@ Partition[Range[n], 2] ===
  ArrayReshape[Range[n], {n/2, 1, 2}]]

(* True *)

Comparing timings

timing1[n_Integer?Positive] :=
 RepeatedTiming[List /@ Partition[Range[n], 2];][[1]]

timing2[n_Integer?Positive] :=
 Module[{data = Range[n]},
  RepeatedTiming[ArrayReshape[data, {Length[data]/2, 1, 2}];][[1]]]

ListLogLogPlot[{
  {#, timing1[#]} & /@ (10^Range[7]),
  {#, timing2[#]} & /@ (10^Range[7])},
 Joined -> True,
 Frame -> True,
 FrameLabel -> (Style[#, 14, Bold] & /@
    {"number of elements", 
     "timing"}),
 PlotLegends ->
  Placed[{"Partition and Map", ArrayReshape}, {.35, .75}]]

Slower than ArrayReshape but ... there is also

BlockMap

BlockMap[List, Range@18, 2]

{{{1, 2}}, {{3, 4}}, {{5, 6}}, {{7, 8}}, {{9, 10}}, {{11, 12}}, {{13, 14}}, {{15, 16}}, {{17, 18}}}

and the (undocumented) 6-argument form of Partition:

Partition[Range@18, 2, 2, 1, {}, {{##}} &]

{{{1, 2}}, {{3, 4}}, {{5, 6}}, {{7, 8}}, {{9, 10}}, {{11, 12}}, {{13, 14}}, {{15, 16}}, {{17, 18}}}

Possibly what you are looking for:

First @ Partition[{Range[18]}, {1, 2}]

{{{1, 2}}, {{3, 4}}, {{5, 6}}, {{7, 8}}, {{9, 10}},
   {{11, 12}}, {{13, 14}}, {{15, 16}}, {{17, 18}}}

Note the extra { } around Range[18] to make it two dimensional.

Or Partitioning sequentially:

Fold[Partition, Range[18], {2, 1}]