You can do this in linear time.
Claim: I can build a data structure in linear time that lets me check, in constant time, whether an arbitrary rectangle is full of 1's.
Proof: Partial sums; take S[i][j] to be the total number of 1's above and to the left of (i, j). The number of ones in the rectangle between (a,b) and (c,d), provided (a,b) is above and left of (c,d), is S[c][d] + S[a][b] - S[a][d] - S[b][c].
Now it's a simple scan over the array:
size = 1;
For i = 0 to m-size {
For j = 0 to n-size {
If S[i+size][j+size] - S[i][j+size] - S[i+size][j] + S[i][j] == size*size {
size++; j--; continue;
}
}
}
At the end, size is one greater than the largest 1-full square.
