Hint/Approach: A way to extend the definition of Stirling numbers are the "Associated Stirling numbers of the second kind", they are denoted by ${ n\brace k}_{\geq m},$ where this means the number of partitions of $[n]$ into $k$ blocks, each one having at least $m$ elements. So, most likely if you know that the formula involves taking Stirling numbers, you can change that for these associated. You can easily check that they satisfy the following recursion
$${n+1\brace k}_{\geq m} = \sum _{j = m-1}^n\binom{n}{j}{n-j\brace k-1}_{\geq m},$$ by checking the elements that are in the same block as the element $n+1.$
First, since there are $n$ possible places to put any of the $m$ balls, we have $n^m$ equally likely distributions of the balls, and we need only to count how many of these have exactly one ball in at least one bin. Note that to get to $n^m$, we assumed that both the balls and the bins are distinguishable. We may assume that the former are numbered from $1$ to $m$ and the latter from $1$ to $n$. From now on, I'll call a bin with exactly one ball a "singleton".
We might try to count the number of distributions with $k$ singletons as follows. There are $\binom nk$ ways to choose the bins, and $\binom mk$ ways to choose the balls to go in them. They can be placed in the bins in $k!$ ways. The remaining $m-k$ balls can be placed in the remaining $n-k$ bins in $(n-k)^{m-k}$ ways, giving a total of $$ \binom nk\binom mk k!(n-k)^{m-k}$$
Unfortunately, this isn't correct, because one or more of the other $n-k$ bins may get exactly one ball. The principle of inclusion and exclusion proceeds by making the incorrect calculation above, and making successive corrections until it arrives at the correct answer.
Before developing the formula, we must deal with one other question. How large can $k$ be? If $n\geq m$ then each ball can go in its own bin, so $k\leq m$. If $m>n$ then at most $n-1$ bins can be singleton, and the remaining bin contains all the other balls. Then
$$k\leq \phi(m,n):=\cases{
m,&$n\geq m$\\
m-1,&$n<m$
}$$
or more simply, $$\phi(m,n) = m-[n<m]$$ where $[\cdot]$ is the Iverson bracket
Now we start with $k=1$ to get $$ \binom n1\binom m1 1!(n-1)^{m-1}$$
We note however, that any distribution that has two singletons will have been counted twice in the above formula, once for each singleton, so we have to subtract the distributions with two singletons, giving $$\binom n1\binom m1 1!(n-1)^{m-1}- \binom n2\binom m2 2!(n-2)^{m-2}$$ Now a distribution with three singletons has been counted three times (once for each singleton) and subtracted three times (once for each pair of them) so we have to add these distributions back in, and so on.
By the inclusion-exclusion principle, the number of distributions containing at least one singleton is $$\sum_{k=1}^{\phi(m,n)}(-1)^{k+1}\binom nk\binom mk k!(n-k)^{m-k}\tag1$$
To compute the desired probability, just divide the value from $(1)$ by $n^m$.
Obviously, the expression in $(1)$ unattractive. I don't know if it can be simplified. At least you can compute for any value of $m$ and $n$, with a computer at any rate.
Best Answer
What is the chance that all $k$ bins are occupied?
For $1\leq i\leq k$, define $A_i$ to be the event that the $i$th bin stays empty. These are exchangeable events with $P(A_1\cdots A_j)=(1-{j\over k})^n$ and so by inclusion-exclusion, the probability that there are no empty bins is $$P(X=0)=\sum_{j=0}^k (-1)^j {k\choose j}\left(1-{j\over k}\right)^n.$$
Stirling numbers of the second kind can be used to give an alternative solution to the occupancy problem. We can fill all $k$ bins as follows: partition the balls $\{1,2,\dots, n\}$ into $k$ non-empty sets, then assign the bin values $1,2,\dots, k$ to these sets. There are ${n\brace k}$ partitions, and for each partition $k!$ ways to assign the bin values. Thus, $$P(X=0)={{n\brace k}\,k!\over k^n}.$$