A couple of additional variants:

(* SET UP *)
SeedRandom[0]; (* optional, for reproducibility *)
dim = 2^16;
rowIdcs = RandomSample[Range[dim], 1000];(*i*)
colIdcs = RandomSample[Range[dim], 1000];(*j*)
entries = RandomReal[1, 1000];(*vals*)

sa1 = SparseArray[ (* first variant *)
   MapThread[{#1, #2} -> #3 &, {rowIdcs, colIdcs, entries}],
   {dim, dim}];

sa2 = SparseArray[ (* second variant *)
   Thread[Transpose[{rowIdcs, colIdcs}] -> entries],
   {dim, dim}];

sa2 == sa1
(*  True  *)

Some are uncomfortable with # and anonymous (or pure) functions. One can make a utility function on the fly. Here a couple variant, plus an anonymous one with named arguments.

ClearAll[entryRule];
entryRule[{i_, j_, val_}] := {i, j} -> val; (* Note {} in argument *)
sa3 = SparseArray[
   entryRule /@
    Thread[{rowIdcs, colIdcs, entries}],(* = triples *)
   {dim, dim}];

ClearAll[entryRuleL]; (* uses Listability instead of Map/Thread *)
SetAttributes[entryRuleL, Listable];
entryRuleL[i_, j_, val_] := {i, j} -> val; (* No {} in arguments *)
sa4 = SparseArray[entryRuleL[rowIdcs, colIdcs, entries], {dim, dim}];

sa5 = SparseArray[ (* anonymous variant of entryRuleL *)
   Function[{i, j, val}, {i, j} -> val, Listable][
    rowIdcs, colIdcs, entries],
   {dim, dim}];

sa5 == sa4 == sa3 == sa2 == sa1
(*  True  *)

Oh well, now it feels embarrassingly like overkill. :)

Your second little grey information box including

Sparsearray[
 ...
{1,1}->{1,1}->20.8
....

shows something is very wrong for you to get {1,1}->{1,1}->20.8

Maybe I broke something in the way I coded that. I apologize.

Let me try coding that slightly differently

i={1,2,7};j={2,5,3};val={1.3,2.9,3.1};
rules=Map[Most[#]->Last[#]&,Thread[{i,j,val}]];
mat=SparseArray[rules]
MatrixForm[mat]

which I thought should be equivalent and which displays for me

Sparsearray[ Specified Elements 3, Dimensions (7,5)

( 0 1.3 0 0 0 )
( 0 0 0 0 2.9 )
( 0 0 0 0 0 )
( 0 0 0 0 0 )
( 0 0 0 0 0 )
( 0 0 0 0 0 )
( 0 0 3.1 0 0 )

Please try my tiny example and see if you get exactly the same result.

And, again this example is only doing the left hand side of your problem. Hopefully you can see how to use these ideas to do the right hand side of your problem.

If so then start with a tiny tiny subset of your 16.7 million elements and see if that works and if so then gradually work up by small amounts and see if it keeps working for larger and larger lists.

Thank you for your help. All is good statements work thank you.