$\require{AMScd}$
The following result holds in much more generality than asked by the OP. There is no characteristic zero/smoothness assumption.

**Proposition:** Let $X/k$ be a qcqs scheme with $k$ separably closed. Let $K/k$ be a separable extension with $K$ separably closed. Let $n \in \mathbf{N}$ be a positive integer that is prime to the characteristic of $k$. Then the natural map
$$\frac{\operatorname{Pic}(X)}{n\operatorname{Pic}(X)} \to \frac{\operatorname{Pic}(X_K)}{n\operatorname{Pic}(X_K)}$$
is an isomorphism.

The proof goes as follows. The Kummer sequence gives the following commutative diagram with exact rows:
$\require{AMScd}$
\begin{CD}
0 @>{}>>\frac{\operatorname{Pic}(X)}{n\operatorname{Pic}(X)} @>{}>> H^2(X, \mu_n)@>{}>> H^2(X,\mathbf{G}_m) \\
{} @VVV @VVV @VVV {} \\
0 @>{}>> \frac{\operatorname{Pic}(X_K)}{n\operatorname{Pic}(X_K)} @>{}>> H^2(X_K, \mu_n) @>{}>> H^2(X_K, \mathbf{G}_m)
\end{CD}

The middle vertical arrow is an isomorphism by the smooth base change theorem in \'{e}tale cohomology (Corollary 1.3.1 of this):

**Smooth Base Change:** Suppose $K/k$ is an extension of separably closed fields, $X$ a qcqs $k$-scheme and $\mathscr{F}$ a torsion sheaf with torsion orders not divisible by the characteristic of $k$.Then, the base change map
$$H^i(X,\mathscr{F}) \to H^i(X_K, \mathscr{F}_K)$$
is an isomorphism for all $i \geq 0$.

Therefore to prove the proposition, we reduce to proving the following injectivity statement concerning Brauer groups:

**Proposition:** Let $k \subseteq K$ be a separable extension of fields with $k$ separably closed. Let $X/k$ be a qcqs scheme. Then $H^2(X,\mathbf{G}_m)$ injects into $H^2(X_K, \mathbf{G}_m)$.

To prove this, write $K$ as a direct limit of smooth $k$-algebras. Since cohomology commutes with direct limits for qcqs schemes, if a (cohomological) Brauer class $\alpha$ in $H^2(X,\mathbf{G}_m)$ dies in $H^2(X_K,\mathbf{G}_m)$ it must die in $H^2(X_R, \mathbf{G}_m)$ for some smooth $k$-algebra $R$. By smoothness and using that $k$ is separably closed, $\operatorname{Spec} R$ has a $k$-rational point. Therefore the canonical map $f: X_{R} \to X$ has a section $g$ and so on cohomology the composition
$$H^2(X,\mathbf{G}_m) \stackrel{f^\ast}{\to} H^2(X_R,\mathbf{G}_m) \stackrel{g^\ast}{\to} H^2(X,\mathbf{G}_m)$$ is the identity. We conclude that $\alpha = 0$ as desired.

Finally we remark that in the injectivity statement on Brauer groups, only the ground field is required to be separably closed. There is no requirement on $K$, only that $K/k$ be separable.

**Edit:** The proof also shows that for any \'{e}tale sheaf $\mathscr{F}$ on $X$ such that multiplication by $n$ is surjectve, then $$\frac{H^i(X,\mathscr{F})}{nH^i(X,\mathscr{F})} \to \frac{H^i(X_K,\mathscr{F})}{nH^i(X_K,\mathscr{F})} $$
is an isomorphism for all $i \geq 1$.