If $\mathcal{L}$ is ample, then $i^*\mathcal{L}$ is ample for example by Morphisms, Lemma \ref{morphisms-lemma-pullback-ample-tensor-relatively-ample}. Assume $i^*\mathcal{L}$ is ample. We have to show that $\mathcal{L}$ is ample on $X$. Let $\mathcal{I} \subset \mathcal{O}_X$ be the coherent sheaf of ideals cutting out the closed subscheme $Z$. Since $i(Z) = X$ set theoretically we see that $\mathcal{I}^n = 0$ for some $n$ by Lemma \ref{lemma-power-ideal-kills-sheaf}. Consider the sequence $$ X = Z_n \supset Z_{n - 1} \supset Z_{n - 2} \supset \ldots \supset Z_1 = Z $$ of closed subschemes cut out by $0 = \mathcal{I}^n \subset \mathcal{I}^{n - 1} \subset \ldots \subset \mathcal{I}$. Then each of the closed immersions $Z_i \to Z_{i - 1}$ is defined by a coherent sheaf of ideals of square zero. In this way we reduce to the case that $\mathcal{I}^2 = 0$. \medskip\noindent Consider the short exact sequence $$ 0 \to \mathcal{I} \to \mathcal{O}_X \to i_*\mathcal{O}_Z \to 0 $$ of quasi-coherent $\mathcal{O}_X$-modules. Tensoring with $\mathcal{L}^{\otimes n}$ we obtain short exact sequences \begin{equation} \label{equation-ses} 0 \to \mathcal{I} \otimes_{\mathcal{O}_X} \mathcal{L}^{\otimes n} \to \mathcal{L}^{\otimes n} \to i_*i^*\mathcal{L}^{\otimes n} \to 0 \end{equation} As $\mathcal{I}^2 = 0$, we can use Morphisms, Lemma \ref{morphisms-lemma-i-star-equivalence} to think of $\mathcal{I}$ as a quasi-coherent $\mathcal{O}_Z$-module and then $\mathcal{I} \otimes_{\mathcal{O}_X} \mathcal{L}^{\otimes n} = \mathcal{I} \otimes_{\mathcal{O}_Z} i^*\mathcal{L}^{\otimes n}$ with obvious abuse of notation. Moreover, the cohomology of this sheaf over $Z$ is canonically the same as the cohomology of this sheaf over $X$ (as $i$ is a homeomorphism). \medskip\noindent Let $x \in X$ be a point and denote $z \in Z$ the corresponding point. Because $i^*\mathcal{L}$ is ample there exists an $n$ and a section $s \in \Gamma(Z, i^*\mathcal{L}^{\otimes n})$ with $z \in Z_s$ and with $Z_s$ affine. The obstruction to lifting $s$ to a section of $\mathcal{L}^{\otimes n}$ over $X$ is the boundary $$ \xi = \partial s \in H^1(X, \mathcal{I} \otimes_{\mathcal{O}_X} \mathcal{L}^{\otimes n}) = H^1(Z, \mathcal{I} \otimes_{\mathcal{O}_Z} i^*\mathcal{L}^{\otimes n}) $$ coming from the short exact sequence of sheaves (\ref{equation-ses}). If we replace $s$ by $s^{e + 1}$ then $\xi$ is replaced by $\partial(s^{e + 1}) = (e + 1) s^e \xi$ in $H^1(Z, \mathcal{I} \otimes_{\mathcal{O}_Z} i^*\mathcal{L}^{\otimes (e + 1)n})$ because the boundary map for $$ 0 \to \bigoplus\nolimits_{m \geq 0} \mathcal{I} \otimes_{\mathcal{O}_X} \mathcal{L}^{\otimes m} \to \bigoplus\nolimits_{m \geq 0} \mathcal{L}^{\otimes m} \to \bigoplus\nolimits_{m \geq 0} i_*i^*\mathcal{L}^{\otimes m} \to 0 $$ is a derivation by Cohomology, Lemma \ref{cohomology-lemma-boundary-derivation-over-cup-product}. By Lemma \ref{lemma-section-affine-open-kills-classes} we see that $s^e \xi$ is zero for $e$ large enough. Hence, after replacing $s$ by a power, we can assume $s$ is the image of a section $s' \in \Gamma(X, \mathcal{L}^{\otimes n})$. Then $X_{s'}$ is an open subscheme and $Z_s \to X_{s'}$ is a surjective closed immersion of Noetherian schemes with $Z_s$ affine. Hence $X_s$ is affine by Lemma \ref{lemma-image-affine-finite-morphism-affine-Noetherian} and we conclude that $\mathcal{L}$ is ample.