First we start with basic connectivity and counting deductions, to get to:
Looking in the lower right corner, the 2 on the right edge must go down to prevent a 2x2. Just above it, the 2x2 R6-7C19-20 must have an unshaded square, but nothing can reach it except the 2 in R5C19. Then the unshaded square in R3C17 can only be reached by the 5 in R7C17, resolving most of the right side.
Moving left, the unshaded square R6C6 can only be reached by the 4 in R6C9. Then the 2x2 block R4-5C4-5 is only reachable by the 3 in R3C5, forcing R4C5 to be unshaded. Now looking at the 5 in R4C10, it can have at most 1 square going to the left. Moreover there is a 2x3 block, R6-R7C10-12 that can only be reached by the 5, and in particular R6C11 must be in its area. The grid thus far:
Moving to the left side:
There is another 2x3 area which cannot be covered except by the 8 in the lower left, namely R9-10C6-8, so we must have R10-C67 unshaded in the 8s area. We cannot isolate any squares underneath this area, so we must in fact have the whole row R10C1-7. Looking at the 3 in R9C10, we see that at least one of its cells must go right, as otherwise R9-10C11-12 would be a shaded 2x2. In particular, this forces the 8 area to continue into R10C8, so that R9-10C8-9 will not be entirely shaded. Now R8-9C1-2 must have an unshaded square, which must include at least R8C2.
Moving up for a moment, we see that the 6 in R2C3 cannot go left, since it would block the 2 in the upper left corner, so the 2 in R5C1 must go up. This forces the 4 in R8C3 to continue up into R7C2, completing the lower left. Finally, the 6 cannot cover the R2-3C1-2 2x2, so the 2 in the upper left corner must come down. The grid thus far:
The 2x2 R9-10C12-13 can only be covered by the 3 in R9C10, which must extend right to R9C12. This then completes the 4 in R9C16, forcing it to extend to R10C14, and also completes the 5 in R4C10, since R7C11 must be unshaded. Now, the 2x2 R5-613-14 can only be reached by the 6 in R1C13, which must extend down, and easy logic finishes this section. The grid thus far:
Since the 6 in R2C3 cannot use R2C4, the 2x2 square R1-2C8-9 can only be reached by the 3 in R3C8, and the rest is just fill-in.