Here is an approach via Lagrange inversion.
Let $N$ denote the time of the first repeat, and let $T(z)$ (the ``tree function'') be the formal power series satisfying $T(z)=z\,e^{T(z)}$.
If $F$ is a formal power series the coefficients of $G(z):=F(T(z))$ are given by (Lagrange inversion)
$$[z^0]G(z)=[z^0] F(z) \mbox{ , } [z^k]G(z)=\tfrac{1}{k} [y^{k-1}] F^\prime(y)\,e^{ky} =[y^k](1-y)F(y)\,e^{ky}\mbox{ for } k\geq 1\;.$$
In particular
$$[z^m] T(z)^k=\frac{k}{m} \frac{m^{m-k}}{(m-k)!}\mbox{ and }
\frac{1}{1-T(z)}=\sum_{n\geq 0}\frac{n^{n}}{n!}z^n$$
Using the first relation it is easily seen that the generating function of $N$ may be written as
$$\mathbb{E} t^N=\frac{m!}{m^m} [z^m]\frac{t}{1- tT(z)}$$
Thus the binomial moments $\mathbb{E}{N \choose k}$ of $N$ can be obtained as
$$\mathbb{E}{N \choose k} = \frac{m!}{m^m} [z^m t^k]\frac{1+t}{1- (1+t)T(z)}=\frac{m!}{m^m} [z^m ]\frac{T(x)^{k-1}}{(1- T(z))^{k+1}}$$
Differentiation shows that $z\,T^\prime(z)= \frac{T(z)}{1-T(z)}$.
Therefore
\begin{align*} \mathbb{E}{N \choose 2} &=\frac{m!}{m^m} [z^m ]\frac{T(z)}{(1- T(z))^{3}}\\
&=\frac{m!}{m^m} [z^{m-1} ]\frac{T^\prime(z)}{(1- T(z))^{2}}\\
&=\frac{m!}{m^m} [z^{m-1} ]\big(\frac{1}{1- T(z)}\big)^\prime\\
&=\frac{m!}{m^m} m\,[z^{m} ]\frac{1}{1- T(z)}
=m\end{align*}
(I don't know who first observed that.)
Similarly
\begin{align*} \mathbb{E}{N \choose 4} &=\frac{m!}{m^m} [z^m ]\frac{T(z)^3}{(1- T(z))^{5}}\\
&=\frac{m!}{m^m} [z^{m-1} ] T^\prime(z)\frac{T(z)^2}{(1- T(z))^{4}}\\
&=\frac{m!}{m^m} [z^{m-1} ]\big(\frac{1}{1- T(z)}-\frac{2}{3}\frac{1}{(1-T(z))^2}+\frac{1}{3}\frac{T(z)}{(1-T(z))^3}\big)^\prime\\
&=\frac{m!}{m^m} m\,[z^{m}]\big(\frac{1}{1- T(z)}-\frac{2}{3}\frac{1}{(1-T(z))^2}+\frac{1}{3}\frac{T(z)}{(1-T(z))^3}\big)\\
&=\frac{m^2}{3} +m -\frac{2}{3} m\,\mathbb{E}(N)\end{align*}
Clearly other binomial moments can be treated in the same way.