So first of all,
a bunch of black squares can be placed, since they're currently dead ends. Using those cells and the 3 clue on the left, we can get this far:
From the top right,
the 2 clue is constrained. You can't have both shaded cells in the top region, meaning the one in row 8 of that column must be shaded. (And at the same time, the 1→ and 2← pointing at each other mean that there can only be one more shaded cell in row 4, and it must be on the left side.)
And now it's time for the main break-in to the puzzle: there's an interesting question hiding in this clue layout.
What happens in the center, exactly?
The board is divided into four "regions", each with only a few ways out.
If a cell is shaded in the center...
we run into a problem. Here, the red and blue regions both need to exit both up and down to be part of the loop -- which means that the yellow and green regions have three endpoints each, and we can't connect those three without leaving a loose end.
There's another question in there, with a similar method.
What is the general placement of shaded cells in each region?
It seems that if we checkerboard-color each region like this:
each region is entered and exited on a dark square. This means that each region has one more dark shaded cell than light shaded cells. And this means that each region has an odd number of shaded cells.
So, armed with this new constraint, we can make some more deductions:
There's no way to fit three shaded cells in the bottom left or top right regions. So the remainder of the bottom left must be empty, and the top right's shaded cell must be in row 5.
More connectivity constraints give more deductions, leading to the solution.
The row-4 shaded cell cannot be in column 4, because it would then provide too many entrances/exits to that region. It cannot be in column 5, because then the shaded cell from the row above that would be forced into column 5 as well. So it must be in column 3. And this leads to the answer: