Assume that n is nonzero.
The probability is the number of permutations with at least one run of m or more heads ("qualifying permutations"), divided by the total number of permutations.
The total number of permutations is 2^(m+n).
The length of runs of heads in qualifying permutations ranges from m to m+n. Because m > n there is no qualifying permutation with two sets of m or longer runs separated by one or more tails. Each qualifying permutation therefore has just one qualifying run of heads, of length between m and m+ n (inclusive).
It's easy to see that there is just one qualifying permutation of length m+n, two of length m+n-1, and if applicable, three of length m+n-2, etc.
The number of qualifying permutations depends on the size of n.
If n =1 there are 1+2 = 3 qualifying permutations.
If n = 2 there are 1+2+3 = 6 qualifying permutations.
If n = 3 there are 1+2+3+4 = 10 qualifying permutations.
And so on, with this arithmetic progression.
Again, probability = qualifying permutations / total permutations.