I realize this puzzle is long forgotten by this time, but I'm slowly working through the archives and consulting the forum for possible posts on any puzzles I've found interesting or more difficult than usual. Like some others, I found this one more time consuming than most Nightmares, so I'm happy to see that I was in good company. I did not find the template eliminations that Ruud explained or the approach that David and Laura used. For what it's worth, here's how I proceeded from the bottleneck position Ruud posted above (repeated below for convenience).
I observed the conjugate pair r67c4 for candidate 3 in column 4. It's clear that setting r6c4 to 3 has some immediate implications, in view of the bivalued peer cell r6c2. It's also clear that setting r7c4 to 1 or 2 has some immediate implications, in view of the bivalued peer cells r3c4 and r7c2, respectively. So further investigation looked promising.
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.------------------.------------------.------------------.
| 23 46 1 | 2789 2689 46789| 4689 3469 5 |
| 23 8 46 | 129 1269 5 | 469 1349 7 |
| 9 5 7 | 18 3 1468 | 2 146 1468 |
:------------------+------------------+------------------:
| 57 123 9 | 4 1568 1368 | 68 2567 268 |
| 6 124 248 | 1589 7 189 | 489 2459 3 |
| 57 34 48 | 3589 5689 2 | 1 45679 468 |
:------------------+------------------+------------------:
| 8 26 5 | 123 4 13 | 7 126 9 |
| 4 79 3 | 6 129 179 | 5 8 12 |
| 1 79 26 | 5789 589 789 | 3 246 246 |
'------------------'------------------'------------------'
Placing 2 in r7c4 would lead to a conflict since
r7c4=2 ==> r7c2=6 ==> r1c2=4 ==> r6c2=3 ==> r6c4 <> 3
leaving column 4 without a 3.
So 2 can be removed from r7c4, creating a naked pair of {1,3} in r7c46, which (among other eliminations) eliminates 1 from r8c6. Now r9c6 must equal 8 in order to avoid the non-unique rectangle which Ruud mentioned.
From the position above (independent of the previously mentioned eliminations) it can also be shown that candidate 1 can be eliminated from r7c4. Setting r7c4=1 would lead to a conflict of two "6" digits in box 6:
r7c4=1 ==> r6c4=3 ==> r6c2=4 ==> r6c3=8 ==> r6c9=6
Also, (making use of the conjugate pair in box 3 for candidate 8)
r7c4=1 ==> r3c4=8 ==> r3c9 <> 8 ==> r1c7=8 ==> r4c7=6.
So we conclude that r7c4=3, and it's straightforward from there, with nothing more advanced than naked pairs needed.