I think the conjecture is true.
Below is an outline of a proof strategy---however, carefully verification of the details remains. If I find more time, I can try to fill those in (or maybe someone else provides a different proof before that).
For $\{A_n\}$ to be a Stieltjes-Moment sequence, two matrices $\Delta$ and $\Delta'$ must be positive definite.
The matrix $\Delta$ is defined as \begin{equation*} \Delta := \begin{pmatrix} A_0 & A_1 & \cdots & A_n\\ A_1 & A_2 & \cdots & A_{n+1}\\ & \vdots & & \\ A_n & A_{n+1} & \cdots & A_{2n} \end{pmatrix},\quad\text{i.e.}\quad \Delta_{ij} = A_{i+j-2}, 1\le i,j \le n+1, \end{equation*} while the matrix $\Delta' := [\Delta'_{ij}] = [A_{i+j-1}]$ for $1 \le i,j \le n+1$.
We prove below that $\Delta$ is symmetric positive semidefinite (a brief additional argument should establish strict positivity, which is what is needed to ensure infinite support).
First, we write $A_n$ using slightly different notation: \begin{equation*} A_n = \sum_{k=0}^n a_{n,k}^2,\qquad a_{n,k} := \binom{n}{k}\binom{n+k}{k}. \end{equation*} Next, define the order-0 Schmidt numbers \begin{equation*} S_n := \sum_{k=0}^n a_{n,k}, \end{equation*} and consider the matrix $M$ formed like $\Delta$ except that instead of $A_n$ we use $S_n$. We begin by proving that $S_n$ is positive definite, in particular by showing that \begin{equation*} S_{i+j-2} = \langle \phi(i), \phi(j) \rangle, \end{equation*} for some $\phi$. A similar (though more involved) technique can be followed for $A_n$ (though, if we actually could represent $a_{i+j-2,k}$ as an inner product, then the proof for $A_n$ would follow immediately using the Schur-product theorem).
The key trick is to use the ``symmetric'' form of the Vandermonde-Chu identity: \begin{equation*} \binom{r+s}{k} = \sum_{p,q \ge 0; p+q=k}\binom{r}{p}\binom{s}{q}. \end{equation*} Applying this identity, we have \begin{eqnarray*} \binom{i+j-2}{k} &=& \sum_{p,q\ge 0, p+q=k} \binom{i-1}{p}\binom{j-1}{q}\\ \binom{i+j-2+k}{k} &=& \sum_{p,q\ge 0, p+q=k} \binom{i-1+k/2}{p}\binom{j-1+k/2}{q}. \end{eqnarray*} Since $\binom{n}{j} = 0$ for $j > n$, we drop the summation indices (unless needed), and obtain \begin{eqnarray*} M_{ij} &=& \sum_{k}\biggl( \sum_{\substack{p,q \ge 0\\ p+q=k}} \binom{i-1}{p}\binom{j-1}{q} \biggr) \biggl( \sum_{p,q \ge 0, p+q=k} \binom{i-1+k/2}{p}\binom{j-1+k/2}{q}\biggr)\\ &=& \sum_k\sum_{\substack{p,q\ge 0, p+q=k\\ r,s \ge0, r+s=k}}\binom{i-1}{p}\binom{i-1+k/2}{r}\binom{j-1}{q}\binom{j-1+k/2}{s}\\ &=& \sum_{p,r,q, s \ge 0}\psi(i; p,r) \psi(j; q,s),\\ &=& \langle \psi(i), \psi(j) \rangle, \end{eqnarray*} for suitably defined $\psi(i; p, r)$. This proves that $M$ is a Gram matrix, hence positive definite.
In a similar way, we can prove that $\Delta_{ij} = \sum_k a_{i+j-2,k}^2 = \langle \phi(i), \phi(j)\rangle$ for a suitable mapping $\phi$, thus establishing positive definiteness of $\Delta$.
Continuing along this path, we can similarly prove $\Delta'$ is also positive definite, which will then finally establish the conjecture.