Let's focus on $P_N:=\sum_{k+j\leq N}\frac{N^{k+j}}{k!j!}\binom{N-j}ku^kv^j$. To get $Z_N$, just multiply out by $\frac{N!}{N^{N+1}}$.

The notation $[z^m]F(z)$ means the coefficient $a_m$ of $z^m$ in the series expansion $F(z)=\sum_na_nz^n$.

Start by changing variables $m=k+j$ so that $j=m-k$ and hence
\begin{align} P_N&=\sum_{m=0}^N\sum_{k=0}^m\frac{N^m}{k!(m-k)!}\binom{N-m+k}ku^kv^{m-k} \\
&=\sum_{m=0}^N\frac{N^m}{m!}\sum_{k=0}^m\binom{m}{m-k}v^{m-k}\binom{N-m+k}ku^k \\
&=\sum_{m=0}^N\frac{N^m}{m!}\cdot[z^m]\left(\frac{(1+vz)^m}{(1-uz)^{N-m+1}}\right) \\
&=\frac1{2\pi i}\sum_{m=0}^N\frac{N^m}{m!}\int_{\gamma}\frac{(1+vz)^m}{(1-uz)^{N-m+1}}\frac{dz}{z^{m+1}};
\end{align}
where in the last line we invoked Cauchy's Integral Formula along a closed path $\gamma$ about $z=0$.

I don't think we can expect a closed formula.