The relationship between the dissociation energy of $N_2$ and $N_2^{+}$ is

The dissociation energy will be more when the bond order will be greater and bond order $\propto$ dissociation energy
Molecular orbital configuration of $N_2\left(14\right)=\sigma 1s^2,\stackrel{\star }{\sigma }1s^2,\sigma 2s^2,\stackrel{\star }{\sigma }2s^2,\sigma 2p_y^2\approx \pi 2p_z^2,\pi 2p_x^2$
So, bond order of $N_2=\frac{N_b-N_a}{2}=\frac{10-4}{2}=3$
and bond order of $N_2^{+}=\frac{9-4}{2}=2.5$
As the bond order of $N_2$ is greater than $N_2^{+}$ so , the dissociation energy of $N_2$ will be greater than $N_2^{+}$