Neat puzzle! The answer is:
The first thing to notice is that the corners with 1s must have their horizontally and vertically adjacent squares shaded. While shading the diagonal is a possibility, it is not required, as we'll see in a moment. The next deduction is to look at the 2s. In particular, the one close to the border in the lower right must have its other empty square be on the border. Examining the possibilities like so
we see that the other configurations cannot have their isolated shaded cells on the border in an L-tromino. Using pink to indicate squares which cannot be shaded, this leaves us with:
Now looking at the other 2:
A similar case analysis shows that three orientations create a shaded region larger than an L-tromino, as below:
Including this, and applying other simple logic to avoid L-tromino adjacencies, we get to this point:
Now look at the lower left corner.
If the box diagonally above and right of the corner were filled, this L-tromino would be surrounded by unshaded squares, except its top corner, which would then have to be shaded to ensure connectivity. This would force a too large shaded piece where it starts to interact with the border around the middle 2, like so:
So we lay two L-trominos in this corner and ensure there is no additional connectivity. But this allows us to define our 13 region, since there are 13 forced unshaded squares, allowing us to finish up some more:
Now look at the 8 region:
There are already 6 unshaded squares in its region. The square down and to the left of the 8 must not be shaded, since there is no way an L-tromino can fit down into it without shading the 8. Moreover, the square to the left of the 8 must be shaded, for if it were not, we would need all four squares above the resulting group of 8 to be shaded, which cannot be done with non-horizontally abutting L-trominos. This forces the following configuration (applying some additional basic logic):
We cannot put an L-tromino around the 9, which means the leftmost 3 squares in the top row, 4 squares in the second row, and 2 squares in the third row must be unshaded. This totals 9, so we must close off this area. Finally, we look at the 1 in the upper left corner, and see that if we do not shade the square diagonally down and left of the one, we cannot make a connected pattern. The rest of the diagram comes together.