Balls and Bins analysis in the m >> n regime.
Not really a full answer (nor a useful reference), but just a rather an extended comment. For any given bin, the probability of having exactly $B$ balls in the bin will be given by $p_B = \binom{m}{B} \left(\frac{1}{n}\right)^B \left(\frac{n-1}{n}\right)^{m-B}$. We can use an inequality due to Sondow, $\binom{(b+1)a}{a}&lt;\left(\frac{(b+1)^{b+1}}{b^b}\right)^a$, to yield $p_B &lt; \left(\frac{(r+1)^{r+1}}{r^r}\right)^B \left(\frac{1}{n}\right)^B \left(\frac{n-1}{n}\right)^{m-B}$, where $r=\frac{m}{B}-1$. Note that this bound is fairly tight, since a $\binom{(b+1)a}{a}&gt;\frac{1}{4ab}\left(\frac{(b+1)^{b+1}}{b^b}\right)^a$.

Thus we have $p_B &lt; e^{B(r+1)\ln(r+1) - Br\ln r - m\ln n + (m-B)\ln (n-1)}$. Now, since you are interested in the probability of finding $B$ or more balls in a bin we can consider $p_{\geq B} = \sum_{b=B}^{m} p_b &lt; \sum_{b=B}^{m} e^{b(r+1)\ln(r+1) - br\ln r - m\ln n + (m-b)\ln (n-1)}$. Rearranging the terms, we get $$p_{\geq B} &lt; e^{-m\ln \frac{n}{n-1}} \times e^{B(r+1)\ln(r+1) - Br\ln r - B\ln (n-1)} \sum_{b=0}^{m-B} e^{b(r+1)\ln(r+1) - br\ln r - b\ln (n-1)}.$$

Note the summation above is merely a geometric series, so we can simplify this to give $$p_{\geq B} &lt; e^{-m\ln \frac{n}{n-1}} \times e^{B(r+1)\ln(r+1) - Br\ln r - B\ln (n-1)} \times \frac{1-\left(\frac{(r+1)^{r+1}}{r^r (n-1)}\right)^{m-B+1}}{1-\left(\frac{(r+1)^{r+1}}{r^r (n-1)}\right)}.$$ If we rewrite $\frac{(r+1)^{r+1}}{r^r (n-1)}$ terms using exponentials, we get 
$$p_{\geq B} &lt; e^{-m\ln \frac{n}{n-1}} \times e^{B(r+1)\ln(r+1) - Br\ln r - B\ln (n-1)} \times \frac{1-\left(e^{(r+1)\ln (r+1) - r \ln r - \ln(n-1)}\right)^{m-B+1}}{1-e^{(r+1)\ln (r+1) - r \ln r - \ln(n-1)}},$$ which then becomes $$p_{\geq B} &lt; \frac{e^{-m\ln \frac{n}{n-1}} \times \left(e^{B((r+1)\ln(r+1) - r\ln r - \ln (n-1))} -e^{(m+1)((r+1)\ln (r+1) - r \ln r - \ln(n-1))}\right)}{1-e^{(r+1)\ln (r+1) - r \ln r - \ln(n-1)}}.$$

Now, I take it you care about finding some $B$ such that $p_{\geq B} &lt; \frac{C}{n}$ for some constant $C$, since this gives the total probability of any bin having $B$ or more balls as bounded from above by $C$. This criteria is satisfied by taking $$\frac{e^{-m\ln \frac{n}{n-1}} \times \left(e^{B((r+1)\ln(r+1) - r\ln r - \ln (n-1))} -e^{(m+1)((r+1)\ln (r+1) - r \ln r - \ln(n-1))}\right)}{1-e^{(r+1)\ln (r+1) - r \ln r - \ln(n-1)}} = \frac{C}{n},$$ which can be rewritten as $$B = \frac{\ln\left(\frac{C}{n} e^{m\ln \frac{n}{n-1}} \left(1-e^{(r+1)\ln (r+1) - r \ln r - \ln(n-1)}\right) + e^{(m+1)((r+1)\ln (r+1) - r \ln r - \ln(n-1))}\right)}{(r+1)\ln(r+1) - r\ln r - \ln (n-1)}.$$

I'm not entirely sure how useful this comment will be to you (it's entirely possible I've made a mistake somewhere), but hopefully it can be of some use.