Let $n$ be a positive integer. Define $$\textbf{A}_n(x):= \left[\frac{1}{x+i+j-1}\right]_{i,j\in\{1,2,\ldots,n\}}$$ as a matrix over the field $\mathbb{Q}(x)$ of rational functions over $\mathbb{Q}$ in variable $x$.
(a) Prove that the Hilbert matrix $\textbf{A}_n(0)$ is an invertible matrix over $\mathbb{Q}$ and all entries of the inverse of $\textbf{A}_n(0)$ are integers.
(b) Determine the greatest common divisor (over $\mathbb{Z}$) of all the entries of $\big(\textbf{A}_n(0)\big)^{-1}$.
(c) Show that $\textbf{A}_n(x)$ is an invertible matrix over $\mathbb{Q}(x)$ and every entry of the inverse of $\textbf{A}_n(x)$ is a polynomial in $x$.
(d) Prove that $x+n$ is the greatest common divisor (over $\mathbb{Q}[x]$) of all the entries of $\big(\textbf{A}_n(x)\big)^{-1}$.
Parts (a) and (c) are known. Parts (b) and (d) are open. Now, Part (d) is known (see i707107's solution below), but Part (b) remains open, although it seems like the answer is $n$.
Recall that $$\binom{t}{r}=\frac{t(t-1)(t-2)\cdots(t-r+1)}{r!}$$ for all $t\in\mathbb{Q}(x)$ and $r=0,1,2,\ldots$. According to i707107, the $(i,j)$-entry of $\big(\textbf{A}_n(x)\big)^{-1}$ is given by $$\alpha_{i,j}(x)=(-1)^{i+j}\,(x+n)\,\binom{x+n+i-1}{i-1}\,\binom{x+n-1}{n-j}\,\binom{x+n+j-1}{n-i}\,\binom{x+i+j-2}{j-1}\,.\tag{*}$$ This means that, for all integers $k$ such that $k\notin\{-1,-2,\ldots,-2n+1\}$, the entries of $\big(\textbf{A}_n(k)\big)^{-1}$ are integers. I now have a new conjecture, which is the primary target for the bounty award.
Conjecture: The greatest common divisor $\gamma_n(k)$ over $\mathbb{Z}$ of the entries of $\big(\textbf{A}_n(k)\big)^{-1}$, where $k$ is an integer not belonging in the set $\{-1,-2,\ldots,-2n+1\}$, is given by $$\gamma_n(k)=\mathrm{lcm}(n,n+k)\,.$$
It is clear from (*) that $n+k$ must divide $\gamma_n(k)$. However, it is not yet clear to me why $n$ should divide $\gamma_n(k)$. I would like to have a proof of this conjecture, or at least a proof that $n \mid \gamma_n(k)$.
Let $M_n$ denote the (unitary) cyclic $\mathbb{Z}[x]$-module generated by $\dfrac{1}{\big((n-1)!\big)^2}\,(x+n)$. Then, the (unitary) $\mathbb{Z}[x]$-module $N_n$ generated by the entries of $\big(\textbf{A}_n(x)\big)^{-1}$ is a $\mathbb{Z}[x]$-submodule of $M_n$.
We also denote by $\tilde{M}_n$ for the (unitary) $\mathbb{Z}$-module generated by $\dfrac{1}{\big((n-1)!\big)^2}\,(x+n)\,x^l$ for $l=0,1,2,\ldots,2n-2$. Then, the (unitary) $\mathbb{Z}$-module $\tilde{N}_n$ generated by the entries of $\big(\textbf{A}_n(x)\big)^{-1}$ is a $\mathbb{Z}$-submodule of $\tilde{M}_n$.
For example, $M_2/N_2$ is isomorphic to the (unitary) $\mathbb{Z}[x]$-module $\mathbb{Z}/2\mathbb{Z}$ (in which $x$ acts trivially), and $\tilde{M}_2/\tilde{N}_2$ is isomorphic to the (unitary) $\mathbb{Z}$-module $\mathbb{Z}/2\mathbb{Z}$. Hence, $\left|M_2/N_2\right|=2=\left|\tilde{M}_2/\tilde{N}_2\right|$. For $n=3$, Mathematica yields $$\tilde{M}_3/\tilde{N}_3\cong (\mathbb{Z}/2\mathbb{Z})\oplus(\mathbb{Z}/3\mathbb{Z})^{\oplus 2}\oplus(\mathbb{Z}/4\mathbb{Z})^{\oplus 3}\,,$$ as abelian groups. That is, $\left|\tilde{M}_3/\tilde{N}_3\right|=1152$. On the other hand, $$M_3/N_3\cong \mathbb{Z}[x] \big/\left(12,2x^2+6x+4,x^4-x^2\right)$$ as $\mathbb{Z}[x]$-modules, which gives $\left|M_3/N_3\right|=576$.
Question: Describe the factor $\mathbb{Z}[x]$-module $M_n/N_n$ and the factor $\mathbb{Z}$-module $\tilde{M}_n/\tilde{N}_n$. It is easily seen that $\left|M_n/N_n\right|\leq\left|\tilde{M}_n/\tilde{N}_n\right|$. What are $\left|M_n/N_n\right|$ and $\left|\tilde{M}_n/\tilde{N}_n\right|$? It can be shown also that the ratio $\dfrac{\left|\tilde{M}_n/\tilde{N}_n\right|}{\left|M_n/N_n\right|}$ is an integer, provided that $\left|\tilde{M}_n/\tilde{N}_n\right|$ is finite. Compute $\dfrac{\left|\tilde{M}_n/\tilde{N}_n\right|}{\left|M_n/N_n\right|}$ for all integers $n>0$ such that $\left|\tilde{M}_n/\tilde{N}_n\right|<\infty$. Is it always the case that $\left|\tilde{M}_n/\tilde{N}_n\right|$ is finite?
Apart from the conjecture above, this question is also eligible for the bounty award. I have not yet fully tried to deal with any case involving $n>3$. However, for $n=4$, the module $\tilde{M}_4/\tilde{N}_4$ is huge: $$ \tilde{M}_4/\tilde{N}_4\cong (\mathbb{Z}/2\mathbb{Z})^{\oplus 2}\oplus(\mathbb{Z}/3\mathbb{Z})^{\oplus 3}\oplus(\mathbb{Z}/8\mathbb{Z})^{\oplus 2}\oplus(\mathbb{Z}/9\mathbb{Z})^{\oplus 2}\oplus(\mathbb{Z}/16\mathbb{Z})\oplus(\mathbb{Z}/27\mathbb{Z})$$ as abelian groups.