First thing I noticed:
Assuming all correct so far (I've not re-done the puzzle, just looking at current state), the second to bottom row could be:
Either way there's an extra 'x' that can be marked (in the column with 1 7 1)
That column has a 7 above a 1, and the extra 'x' means the block for the 1 can be placed, and more blocks from the 7 become certain...
Second thing, and perhaps more useful:
Looking at it on its own, the '5' at the far right can be in two different distinct positions...
However, if it were in the lower of the 2 positions, that immediately creates a contradiction, as it forces more than 3 consecutive cells to be highlighted in the second-to-right column. Therefore it must be in the top half of the grid.
This directly implies a stronger form of the first thing I noticed, and several other cascading implications.
Following through with that I was able to complete the puzzle yielding the following grid:
So it seems that nobody made a mistake - your solution so far was correct, and the puzzle was solvable.
Solving using only "simple logic"
The column with 1 7 1 forces an additional square to be shaded - you perhaps mis-counted how many squares were forced.
This additional square is one that would have to be blank for the 2 3 2 from its row to all fit in the same wide gap, therefore that row must have the final 2 at the far right.
That then indicates which gap the '5' fits into without calling on the logic given above, and the rest of the puzzle will fall into place as before.