I am really trying not to guess, but have been stumped for a while here. I have cleared out a few impossible, but they weren't enough. Anyone can explain the next move?

EDIT: solved below!

Last column you have two 1,6 squares which means 1 and 6 have to go in either R1C9 or R6C9. Therefore neither can go in R5C9 making it an 8

Remove all the 1s and 6s on cells in row 6, except the cells with only 1 and 6

Its only 1 square w/ 1,6 in the column

Last column, 3 squares that contain only 1, 6, or 8. Means you can remove those numbers from the rest of the column, giving you the results of the other 2 squares in that column.

(Also, there's nowhere else your 3 in the final column could go anyway)

You have three cells in the last colmn that only have 1, 8, or 6, and no other nuber. That means that no other cell can have 1, 8 or 6.

So the second and third cell from the top cannot have 1,8 and 1,6 respectively. You can try it and see that it wouldn't work.

Also a bit of a cool one, but on the row just above, you can't have any 8s except in the second or last cell from the left. Or else it would make the sudoku have 2 solutions, which is against the normal rule afaik.

They are not all In the same column, so you can't eliminate it.

There are definitely 3 squares in the last column that can only be 1, 6, or 8, with no other possibilities: R1C9, R5C9, R6C9. That means that R2C9 and R3C9 can't include a 1, 6, or an 8.

Additionally, your 3s are confined to columns 7 and 8 in boxes 6 and 9, which means that you can't possibly have a 3 in R2C8. This makes R2C9 a 3, and therefore R3C9 a 5.

Making these moves opens a few other moves up in the board.

Ahh I follow that now! Thanks, this unlocked the rest

My eyes deceived me, instead, look where 3 can go in the last column