You'll want to prove separately that $\Gamma \vdash_H M$ implies $\Gamma \vdash_N M$ and vice versa. Each of these cases can be done by induction on the structure of the proof tree (or by the length of a proof, if your $H$ represents proofs as flat sequences of formulae rather than trees).
The direction ${\vdash_H} \Rightarrow {\vdash_N}$ is easy, since the only inference rule in $H$ is modus ponens, which is also a rule of $N$ (usually called $\to$-elimination in that setting). So all you really have to do is to show that each (instance of a) logical axiom of $H$ can be proved in $N$.
For the direction ${\vdash_N} \Rightarrow {\vdash_H}$, most of the cases are easy. The inference rules of $N$ that don't add anything to the $\Gamma$s can simply be replaced by local derivations of $H$. For example, for $\land$-introduction
$$ \frac{\Gamma\vdash_N M \qquad \Gamma\vdash_N K}{\Gamma \vdash_N M\land K} \,{\land I} $$
show once and for all that
$$ \vdash_H M \to ( K \to M \land K ) $$
for all $M$ and $K$ -- usually this is immediate because that is an axiom of the Hilbert system, but for some systems a bit of proofwork will be needed. Then every application of $\land I$ can be replaced by this proof and two applications of MP after the premises have been translated to $\vdash_H$ by the induction hypothesis.
The main case in a typical natural deduction system (for propositional calculus) where this will not work is $\to$-introduction. Here, exactly what you need to make the case go through is the deduction theorem for $\vdash_H$, which hopefully you have proved already for its own sake.
The $\lor$-elimination rule in its most common formulation will require a combination of these approaches. Apply the deduction theorem to the two premises that extend $\Gamma$, and then show that
$$ \vdash_H (K\lor L)\to((K\to M)\to((L\to M)\to M)) $$
and apply MP three times.
ND_hilbert_equiv
at the end. $\endgroup$