We first try to simplify the sum
$$P_{m,n}(x) = \sum_{k=0}^m {2x+2k\choose 2k+1}
{n+m-k-x-1/2\choose m-k}.$$
We get
$$- [z^m] (1+z)^{n+m-x-1/2}
\sum_{k\ge 0} {-2x\choose 2k+1} z^k (1+z)^{-k}.$$
Here we have extended to infinity due to the coefficient extractor.
Next,
$$- [z^m] (1+z)^{n+m-x-1/2}
\sum_{k\ge 0} z^k (1+z)^{-k}
[w^{2k+1}] (1+w)^{-2x}
\\ = - [z^{2m}] (1+z^2)^{n+m-x-1/2}
\sum_{k\ge 0} z^{2k} (1+z^2)^{-k}
[w^{2k+1}] (1+w)^{-2x}
\\ = - [z^{2m+1}] (1+z^2)^{n+m-x-1/2} \sqrt{1+z^2}
\sum_{k\ge 0} z^{2k+1} (1+z^2)^{-k-1/2}
[w^{2k+1}] (1+w)^{-2x}
\\ = - [z^{2m+1}] (1+z^2)^{n+m-x}
\frac{1}{2} ((1+z/\sqrt{1+z^2})^{-2x}
-(1-z/\sqrt{1+z^2})^{-2x})
\\ = - [z^{2m+1}] (1+z^2)^{n+m}
\frac{1}{2} ((z+\sqrt{1+z^2})^{-2x}
-(-z+\sqrt{1+z^2})^{-2x}).$$
We know from the initial representation that this is a polynomial in
$x.$ Extract the coefficient on $[x^q]$
$$[x^q] \exp(-2x \log(\pm z+\sqrt{1+z^2}))
= \frac{(-1)^q 2^q}{q!} \log^q(\pm z+\sqrt{1+z^2}).$$
The logarithms are both valid formal power series because the
arguments are formal power series that start with constant coefficient
equal to one. Note however that
$$\log(-z+\sqrt{1+z^2}) = - \log\frac{1}{-z+\sqrt{1+z^2}}
\\ = -\log\frac{z+\sqrt{1+z^2}}{1+z^2-z^2}
= -\log(z+\sqrt{1+z^2}).$$
This yields
$$[x^q] P_{m,n}(x)
= - [z^{2m+1}] (1+z^2)^{n+m}
\frac{(-1)^q 2^q}{q!}
\frac{1}{2} \log^q(z+\sqrt{1+z^2}) (1-(-1)^q).$$
This is zero when $q$ is even (inspect last term) and we have the
claim. We get for $q$ odd the coefficient
$$\frac{2^q}{q!} [z^{2m+1}] (1+z^2)^{n+m}
\log^q(z+\sqrt{1+z^2}).$$
The powered logarithm starts at $z^q$ so we get zero when $q\gt 2m+1.$
To see positivity note that an alternate representation is
$$(-1)^{m+1} [z^m] (1+z)^{x-1/2-n}
\sum_{k\ge 0} {-2x\choose 2k+1} (-1)^k z^k
\\ = (-1)^{m+1} [z^m] (1+z)^{x-1/2-n}
\sum_{k\ge 0} (-1)^k z^k [w^{2k+1}] (1+w)^{-2x}
\\ = (-1)^{m+1} [z^{2m+1}] (1+z^2)^{x-1/2-n}
\sum_{k\ge 0} (-1)^k z^{2k+1} [w^{2k+1}] (1+w)^{-2x}
\\ = i (-1)^m [z^{2m+1}] (1+z^2)^{x-1/2-n}
\sum_{k\ge 0} i^{2k+1} z^{2k+1} [w^{2k+1}] (1+w)^{-2x}
\\ = \frac{i}{2} (-1)^m [z^{2m+1}] (1+z^2)^{x-1/2-n}
((1+iz)^{-2x}-(1-iz)^{-2x})
\\ = \frac{i}{2} (-1)^m [z^{2m+1}] (1+z^2)^{-1/2-n}
\left[ \left( \frac{1-iz}{1+iz} \right)^x
- \left( \frac{1+iz}{1-iz} \right)^x \right].$$
Extract the coefficient on $[x^q]$ to get
$$\frac{i}{2q!} (-1)^m [z^{2m+1}] (1+z^2)^{-1/2-n}
\left[ \log^q \frac{1-iz}{1+iz}
- \log^q \frac{1+iz}{1-iz} \right]
\\ = \frac{i}{2q!} (-1)^m [z^{2m+1}] (1+z^2)^{-1/2-n}
\log^q \frac{1-iz}{1+iz} (1-(-1)^q).$$
Zero again for $q$ even, and for $q$ odd,
$$\frac{i}{q!} (-1)^m [z^{2m+1}] (1+z^2)^{-1/2-n}
\log^q \frac{1-iz}{1+iz}
\\ = \frac{1}{q!} [z^{2m+1}] (1-z^2)^{-1/2-n}
\log^q \frac{1+z}{1-z}.$$
The two terms in the product both have series in positive terms only.
To see this, note that the first one is
$$\sum_{p\ge 0} {-1/2-n\choose p} (-1)^p z^{2p}
= \sum_{p\ge 0} {p+n-1/2\choose p} z^{2p}.$$
The coefficients are positive because $(p+n-1/2)^{\underline{p}}$ is.
Note also that
$$\log\frac{1+z}{1-z}
= \log\left(1+\frac{2z}{1-z}\right)
= \sum_{p\ge 1} (-1)^{p-1} \frac{2^p z^p}{p (1-z)^p}.$$
This means the coefficient on $[z^r]$ is
$$\sum_{p=1}^r (-1)^{p-1} 2^p [z^{r-p}] \frac{1}{p (1-z)^p}
= \sum_{p=1}^r (-1)^{p-1} 2^p {r-1\choose p-1} \frac{1}{p}
\\ = \frac{1}{r} \sum_{p=1}^r (-1)^{p-1} 2^p {r\choose p}
= - \frac{1}{r} ((1-2)^r-1) = - \frac{1}{r} ((-1)^r-1).$$
This is zero when $r$ is even and $\frac{2}{r}$ when $r$ is odd. We
have established the desired positivity of the coefficients, as we
have the product of a series with positive coefficients times a power
of a series that also has positive coefficients.
