It follows at once from the properties of sheafification, which is unique up to isomorphism of sheaves.
Indeed. given any morphism of presheaves $\mathscr{F}\longrightarrow \mathscr{G}$ there exists an unique factorisation $\mathscr{F}\to \mathscr{F}^+\to\mathscr {G}$.
Edit.
Someone in the comments pointed out that you need also the extension property for presheaves defined over a basis, which sometimes are called $\mathcr{B}$-presheaves. Of course in order to apply sheafification you need to deal with an actual sheaf, and I assumed that the user was fine with the extension process.
However, let me expand this. Given a basis $\mathscr{B}$ and the $\mathscr{B}$-presheaf $U\mapsto \mathrm{nil}\,\mathscr{O}_X(U)$, the way is the following:
(Extension to a presheaf over $X$) there is an unique presheaf $\mathscr{N}$ over $X$ such that $\mathscr{N}(U)=\mathrm{nil}\,\mathscr{O}_X(U)$;
(Sheafification) there are a sheaf $\mathscr{N}^+$ over $X$ and a morphism $\eta :\mathscr{N}\rightarrow \mathscr{N}^+$ such that for every sheaf $\mathscr{G}$ over $X$ and for every morphism of presheaves $f:\mathscr{N}\longrightarrow \mathscr{G}$ there is an unique morphism of sheaves $g:\mathscr{N}^+\longrightarrow \mathscr{G}$ such that $f=g\circ \eta$.
In particular, the second property shows that $\mathscr{N}^+$ is unique up to isomorphism. So if you realise that the presehaf $\mathscr{N}$ is an actual sheaf you get the isomorphism $\mathscr{N}\simeq \mathscr{N}^+$ for free.