Picard group of a finite type $\mathbb{Z}$-algebra
This is false. A counterexample is given in [Kahn06, Rmq. 1 (6)]. The example uses the cuspidal cubic $B = A[x^2,x^3]$ over a finite type $\mathbb Z$-algebra $A$ that is not a finitely generated $\mathbb Z$-module. For example, take $A$ to be $\mathbb Z[x]$ or $\mathbb F_p[x]$.
As usual, one shows that (under suitable hypotheses on $A$) there is an isomorphism $$\operatorname{Pic}(B) \cong \mathbb G_a(A) = A,$$ which shows that $\operatorname{Pic}(B)$ is not finitely generated. (Details omitted, unfortunately also in the paper.)
References.
[Kahn06] Kahn, Bruno, Sur le groupe des classes d'un schéma arithmétique, Bull. Soc. Math. Fr. 134, No. 3, 395–415 (2006). ZBL1222.14048.
Here I write out some details in R. van Dobben de Bruyn's answer. This argument is from p. 5 of these notes by Konrad Voelkel.
Let $R$ be a Noetherian normal domain with $\mathrm{Pic}(R) = 0$. Set \begin{align*} A := R[x,y]/(x^{3} = y^{2}) \end{align*} which is the cuspidal cubic over $R$. Then \begin{align*} \operatorname{Pic}(A) \simeq R \end{align*} as abelian groups.
The normalization of $A$ is \begin{align*} B := R[t] \end{align*} via the $R$-algebra map $A \to B$ sending $(x,y) \mapsto (t^{2},t^{3})$ which identifies $A$ with the subring $R[t^{2},t^{3}]$ of $B$. Let \begin{align*} I := \{a \in A \;:\; aB \subset A\} \end{align*} be the conductor ideal of the inclusion $A \subseteq B$; it is the largest ideal of $B$ contained in $A$. We have \begin{align*} I = \langle x,y \rangle A \end{align*} (for this, use that $A,B$ are $\mathbb{Z}_{\ge 0}$-graded rings and that $A \subset B$ is a graded ring map, hence $I$ is also a graded ideal of $A$; certainly $\langle x,y \rangle A \subset I$ and if $a \in A$ is nonzero then $a \not\in I$ since otherwise $at \in A$). We have a Milnor square $\require{AMScd}$ \begin{CD} A @>>> B \\ @VVV @VVV\\ A/I @>>> B/I \end{CD} where $A/I \simeq R$ and $B/I \simeq R[t]/(t^{2})$. By [Wei13, I, Theorem 3.10] we have an exact sequence \begin{align*} (A/I)^{\times} \oplus B^{\times} \stackrel{\alpha}{\to} (B/I)^{\times} \to \operatorname{Pic}(A) \to \operatorname{Pic}(A/I) \oplus \operatorname{Pic}(B) \end{align*} of abelian groups (the Units-Pic sequence). Here $\operatorname{Pic}(A/I) = \operatorname{Pic}(R) = 0$ and $\operatorname{Pic}(B) \simeq \operatorname{Pic}(R) = 0$ where the first isomorphism follows from e.g. Traverso's theorem [Wei13, I, Theorem 3.11]. We have $(A/I)^{\times} = R^{\times}$ and $B^{\times} = R^{\times}$ (since $R$ is reduced) and $(B/I)^{\times} = (R[t]/(t^{2}))^{\times} \simeq R \oplus R^{\times}$ (where the "$R$" in "$R \oplus R^{\times}$" is viewed as an abelian group under addition). Under these identifications, the map $\alpha$ sends $(u_{1},u_{2}) \mapsto (0,u_{1}u_{2}^{-1})$ so $\operatorname{coker} \alpha \simeq R$.
References:
[Wei13] Weibel, The K-book: An Introduction to Algebraic K-theory, volume 145 of Graduate Studies in Mathematics. American Mathematical Society (2013)