Here's a quick proof over a finite field $\mathbb F_q$. Suppose $X$ of dimension $n$ embedded in $\mathbb P^N_{\mathbb F_q} $ has this property. Every hyperplane complement in $\mathbb P^N(\mathbb F_q)$ contains $q^N$ points, of which $q^n$ lie in $X(\mathbb F_q)$. So the fraction of points in the hyperplane complement that also lie in $X(\mathbb F_q) $ is $q^{n-N}$. Averaging over all hyperplane complements, as each point in $\mathbb P^{N}(\mathbb F_q)$ is in the same number of hyperplane complements, we see that the fraction of points in $\mathbb P^N(\mathbb F_q)$ that also lie in $X(\mathbb F_q)$ is $q^{n-N}$. Thus $$|X(\mathbb F_q) |= q^{n-N} |\mathbb P^N(\mathbb F_q)| = q^n + q^{n-1} + \dots q^{n-N}$$ which is not an integer unless $N \leq n$, in which case $X = \mathbb P^n$ because it is an $n$-dimensional subscheme of $\mathbb P^N$.

Over an arbitrary field I believe we can do this by a spreading out argument. First, define a stratification of the dual variety, and, over each stratum, an explicit isomorphism from the family of hyperplane complements to $\mathbb A^n$. To do this, choose an isomorphism at the generic point, which automatically extends to an open set, and induct. Next, choose a finitely generated ring over which $X$, the projective embedding, this stratification, and all these isomorphisms are defined. Over an open subset, the stratification will remain a stratification and the isomorphisms will remain isomorphisms. Choosing a finite field-valued point in that subset and performing a counting argument, we see $n=N$ and conclude $X = \mathbb P^n= \mathbb P^N$.

Unless I've made some mistake, no cohomology is needed.

Here's a quick proof over a finite field $\mathbb F_q$, if we assume that the isomorphisms with $\mathbb A^n$ are defined over $\mathbb F_q$. Suppose $X$ of dimension $n$ embedded in $\mathbb P^N_{\mathbb F_q} $ has this property. Every hyperplane complement in $\mathbb P^N(\mathbb F_q)$ contains $q^N$ points, of which $q^n$ lie in $X(\mathbb F_q)$. So the fraction of points in the hyperplane complement that also lie in $X(\mathbb F_q) $ is $q^{n-N}$. Averaging over all hyperplane complements, as each point in $\mathbb P^{N}(\mathbb F_q)$ is in the same number of hyperplane complements, we see that the fraction of points in $\mathbb P^N(\mathbb F_q)$ that also lie in $X(\mathbb F_q)$ is $q^{n-N}$. Thus $$|X(\mathbb F_q) |= q^{n-N} |\mathbb P^N(\mathbb F_q)| = q^n + q^{n-1} + \dots q^{n-N}$$ which is not an integer unless $N \leq n$, in which case $X = \mathbb P^n$ because it is an $n$-dimensional subscheme of $\mathbb P^N$.

If we replace the assumption by the statement that every hyperplane complement defined over $\mathbb F_q$ has an isomorphism with $\mathbb A^n$ defined over $\overline{\mathbb F}_q$, we can do it with a quick etale cohomology lemma.

Every variety $Y$ over $\mathbb F_q$ with $Y_{\overline{\mathbb F}_q} \cong \mathbb A^n_{\overline{\mathbb F}_q}$ has $Y(\mathbb F_q) = q^n$.

Proof: By the Grothendieck-Lefschetz fixed-point formula and the calculation of the cohomology of affine space we have $$Y(\mathbb F_q) =\sum_{i}(-1)^i \operatorname{tr}(\operatorname{Frob}_q, H^i_c(X, \mathbb Q_{\ell}) = \operatorname{tr} (\operatorname{Frob}_q, H^{2n}_c(\mathbb Q_{\ell})).$$ Furthermore, $H^{2n}_c$ is one-dimensional and the Frobenius eigenvalue on it can be computed by Poincare duality as $q^n$, because the unique Frobenius eigenvalue on $H^0(Y_{\overline{\mathbb F}_q}, \mathbb Q_\ell)$ must be one.

With this enhanced form, we can get the statement over an arbitrary field by a spreading out argument. First, define a stratification of the dual variety, and, over a finite etale covering each stratum, an explicit isomorphism from the family of hyperplane complements to $\mathbb A^n$. To do this, choose an isomorphism at the geometric generic point, which automatically extends to a finite etale cover of an open set, and induct. Next, choose a finitely generated ring over which $X$, the projective embedding, this stratification, the finite etale coverings, and all these isomorphisms are defined. Over an open subset, the stratification will remain a stratification and the isomorphisms will remain isomorphisms. Choosing a finite field-valued point in that subset and performing a counting argument, we see $n=N$ and conclude $X = \mathbb P^n= \mathbb P^N$.

Here's a quick proof over a finite field $\mathbb F_q$. Suppose $X$ of dimension $n$ embedded in $\mathbb P^N_{\mathbb F_q} $ has this property. Then for any point $x \in X(\mathbb F_q) $, the probability that $x$ is contained in a random hyperplane section of $\mathbb P^N_{\mathbb F_q}$ is $\frac{ q^{N}-1}{ q^{N+1} -1}$. So the probability that $x$ is contained in a random hyperplane complement is $ = \frac{ q^{N+1} - q^{N}}{ q^{N+1}-1} = (1-1/q) \frac{q^{N+1}}{ q^{N+1} -1} $.

Hence if every hyperplane complement contains $q^n$ points of $X$, certainly a random hyperplane complement contains $q^n$ points on average, so $$(1-1/q) \frac{q^{N+1}}{ q^{N+1} -1} X(\mathbb F_q) = q^n$$ and thus $$X (\mathbb F_q) = (q^n + q^{n-1} + \dots) (1 - q^{-N-1}) = q^n + q^{n-1} + \dots + q^{n-N}. $$

This isn't an integer unless $N\leq n$, in which case $X = \mathbb P^n$.

Over an arbitrary field I believe we can do this by a spreading out argument. First, define a stratification of the dual variety, and, over each stratum, an explicit isomorphism from the family of hyperplane complements to $\mathbb A^n$. To do this, choose an isomorphism at the generic point, which automatically extends to an open set, and induct. Next, choose a finitely generated ring over which $X$, the projective embedding, this stratification, and all these isomorphisms are defined. Over an open subset, the stratification will remain a stratification and the isomorphisms will remain isomorphisms. Choosing a finite field-valued point in that subset and performing a counting argument, we see $n=N$ and conclude $X = \mathbb P^n= \mathbb P^N$.

Unless I've made some mistake, no cohomology is needed.

Here's a quick proof over a finite field $\mathbb F_q$. Suppose $X$ of dimension $n$ embedded in $\mathbb P^N_{\mathbb F_q} $ has this property. Every hyperplane complement in $\mathbb P^N(\mathbb F_q)$ contains $q^N$ points, of which $q^n$ lie in $X(\mathbb F_q)$. So the fraction of points in the hyperplane complement that also lie in $X(\mathbb F_q) $ is $q^{n-N}$. Averaging over all hyperplane complements, as each point in $\mathbb P^{N}(\mathbb F_q)$ is in the same number of hyperplane complements, we see that the fraction of points in $\mathbb P^N(\mathbb F_q)$ that also lie in $X(\mathbb F_q)$ is $q^{n-N}$. Thus $$|X(\mathbb F_q) |= q^{n-N} |\mathbb P^N(\mathbb F_q)| = q^n + q^{n-1} + \dots q^{n-N}$$ which is not an integer unless $N \leq n$, in which case $X = \mathbb P^n$ because it is an $n$-dimensional subscheme of $\mathbb P^N$.

Over an arbitrary field I believe we can do this by a spreading out argument. First, define a stratification of the dual variety, and, over each stratum, an explicit isomorphism from the family of hyperplane complements to $\mathbb A^n$. To do this, choose an isomorphism at the generic point, which automatically extends to an open set, and induct. Next, choose a finitely generated ring over which $X$, the projective embedding, this stratification, and all these isomorphisms are defined. Over an open subset, the stratification will remain a stratification and the isomorphisms will remain isomorphisms. Choosing a finite field-valued point in that subset and performing a counting argument, we see $n=N$ and conclude $X = \mathbb P^n= \mathbb P^N$.

Unless I've made some mistake, no cohomology is needed.