Basically, there are no special techniques for gaussian elimination in $$$\mathbb{Z}/2\mathbb{Z}$$$, but the speedup from $$$O(n^3)$$$ to $$$O(\frac{n^3}{w})$$$ is quite easy ($$$w$$$ is the size of machine word, which is $$$32$$$ or $$$64$$$, depending on the compiler.
The most "expensive" operation in ordinary gaussian elimination is, basically, summation of two rows: we do $$$n$$$ iterations and do $$$O(n)$$$ summations per one iteration, each summation takes $$$O(n)$$$ time, which leads to $$$O(n^3)$$$ complexity.
The key fact that addition of binary vectors can be done in $$$O(\frac{n}{w})$$$ using std::bitset (a += b is just a ^= b), and one can easily achieve $$$O(\frac{n^3}{w})$$$ complexity by just doing all operations with bitsets instead of arrays/vectors.

Hi there, I was just working on a problem and stumble on this topic guided by a chat with an AI. The AI actually proposed an algorithm that runs in O($$$n^2$$$ $$$log\ n$$$) and performs gaussian elimination treating the rows as numbers (as I guess you usually do in these kinds of problem) and using a minimum between $$$a_i$$$ and $$$a_i$$$ ^ $$$b$$$ for some linearly independent row b.

I'll leave here the code for the solution. I would like to ask if anybody knows how or why it works.

vector<long long> find_basis(const vector<long long>& nums) {
    vector<long long> basis;
    for (long long num : nums) {
        long long x = num;
        for (long long b : basis) {
            x = min(x, x ^ b);
        }
        if (x > 0) {
            basis.push_back(x);
            sort(basis.begin(), basis.end(), greater<long long>());
        }
    }
    return basis;
}