Disclaimer: this is only a partial answer,
but something may be better than nothing.
The approach is a brute force method, programming is in Pascal:
program china_math;
function twee(macht : integer) : integer;
{
Compute 2^macht
}
var
k : integer;
h : integer;
begin
h := 1;
for k := 1 to macht do
h := h shl 1;
twee := h;
end;
procedure test(m : integer; blah : boolean);
{
The problem
}
var
i,j,k,a,b,c,d : integer;
s,N,ii,jj,kk : integer;
OK : boolean;
begin
N := twee(m);
s := 0;
ii := (N div 2);
for i := 1 to ii do
begin
a := 2*i-1;
jj := ((N div a) div 2)+1;
for j := 1 to jj do
begin
b := 2*j-1;
kk := ((N-a*b) div 2)+1;
if (N-a*b) > 0 then
for k := 1 to kk do
begin
c := 2*k-1;
OK := (N-a*b) mod c = 0;
if not OK then Continue;
d := (N-a*b) div c;
if blah then Writeln(a,'*',b,'+',c,'*',d,'=',N);
s := s + a*c;
end;
end;
end;
if not blah then
Writeln(m:6,s:12,' =?= ',twee(3*m-3));
end;
procedure doen;
var
m : integer;
begin
test(4,true);
Writeln;
for m := 1 to 11 do
test(m,false);
end;
begin
doen;
end.
Computational details for e.g. $m=4$:
a*b+c*d=2^m
1*1+1*15=16
1*1+3*5=16
1*1+5*3=16
1*1+15*1=16
1*3+1*13=16
1*3+13*1=16
1*5+1*11=16
1*5+11*1=16
1*7+1*9=16
1*7+3*3=16
1*7+9*1=16
1*9+1*7=16
1*9+7*1=16
1*11+1*5=16
1*11+5*1=16
1*13+1*3=16
1*13+3*1=16
1*15+1*1=16
3*1+1*13=16
3*1+13*1=16
3*3+1*7=16
3*3+7*1=16
3*5+1*1=16
5*1+1*11=16
5*1+11*1=16
5*3+1*1=16
7*1+1*9=16
7*1+3*3=16
7*1+9*1=16
9*1+1*7=16
9*1+7*1=16
11*1+1*5=16
11*1+5*1=16
13*1+1*3=16
13*1+3*1=16
15*1+1*1=16
Output for $1 \le m \le 11$:
m sum =?= 2^(3*m-3)
1 1 =?= 1
2 8 =?= 8
3 64 =?= 64
4 512 =?= 512
5 4096 =?= 4096
6 32768 =?= 32768
7 262144 =?= 262144
8 2097152 =?= 2097152
9 16777216 =?= 16777216
10 134217728 =?= 134217728
11 1073741824 =?= 1073741824
This leads to the following Conjecture:
$$\sum_{\substack{ab+cd=2^m\\ a,b,c,d \text{ are odd}}}ac \,=\, 2^{\,3m-3}$$
And the numerical work shows that the Conjecture is a Theorem for $1 \le m \le 11$ ;
which is as far as we can go with the given limited precision of standard (Delphi) Pascal.