I think the solution would go something like this:

Let $P(m, n, k, x, y)$ be the probability of seeing exactly $x$ cards at least once over $k$ draws, once $y$ cards have already been seen.

Then $P(m, n, k, x, 0) = \sum_{i=0}^{x}P(m, n, k-1, x-i, 0) * P(m, n, 1, i, x-i)$

I think the possibilities under the sum sign are all unique, so we can sum them.

Your solution is then $P(m, n, k, m, 0)$. What is left is to define $P(m, n, 1, i, x-i)$ and we should be able to solve the problem recursively.

Edit: full solution in Python, implementing the method above:
<!-- language: lang-py -->

    import numpy as np
    from scipy.special import comb
    import matplotlib.pyplot as plt
    m = 10
    n = 4
    def P(k, x, y):
        if k == 1:
            return (comb(m-y, x) * comb(y, n-x))/comb(m, n)
        else:
            prob = 0
            for i in range(x):
                prob += P(k-1, x-i, y) * P(1, i, y+x-i)
            return prob
    
    def P_MC(k, x, y):
        sims = 10000
        good = 0
        for s in range(sims):
            ar = np.arange(m)
            seen = set(np.arange(y))
            for draw in range(k):
                np.random.shuffle(ar)
                for el in ar[:n]:
                    seen.add(el)
            if len(seen) == (x+y):
                good += 1
        return good/sims

    ests = []
    acts = []
    for k in range(1,16):
        ests.append(P_MC(k, m, 0))
        acts.append(P(k, m, 0))
        
    plt.plot(range(1,16), ests)
    plt.plot(range(1,16), acts)
    plt.grid()
    plt.legend(['Simulated', 'Actual'], loc='lower right')