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.