Starting from (the contribution from $k=0$ is zero owing to the third
binomial coefficient)
$$\sum_{k=1}^n \left(-\frac{1}{4}\right)^k
{2k\choose k}^2 \frac{1}{1-2k} {n+k-2\choose 2k-2}$$
we seek to show that this is zero when $n\gt 1$ is odd and
$$\left[\left(\frac{1}{4}\right)^m
{2m\choose m} \frac{1}{1-2m}\right]^2$$
when $n=2m$ is even.
We observe that with $k\ge 1$
$${2k\choose k} \frac{1}{1-2k} {n+k-2\choose 2k-2}
= 2 {2k-1\choose k-1} \frac{1}{1-2k} {n+k-2\choose 2k-2}
\\ = -2 {2k-2\choose k-1} \frac{1}{k} {n+k-2\choose 2k-2}
= -\frac{2}{k} \frac{(n+k-2)!}{(k-1)!^2 \times (n-k)!}
\\ = -\frac{2}{k} {n+k-2\choose k-1} {n-1\choose k-1}
= -\frac{2}{n} {n\choose k} {n+k-2\choose k-1}.$$
We get for our sum
$$-\frac{2}{n} \sum_{k=1}^n
{n\choose k} \left(-\frac{1}{4}\right)^k
{2k\choose k}
{n+k-2\choose k-1}
\\ = -\frac{2}{n} \sum_{k=1}^n
{n\choose k} {-1/2\choose k}
{n+k-2\choose n-1}
\\ = -\frac{2}{n} [z^{n-1}] (1+z)^{n-2} \sum_{k=1}^n
{n\choose k} {-1/2\choose k} (1+z)^k.$$
The value $k=0$ contributes zero:
$$-\frac{2}{n} \times
\;\underset{w}{\mathrm{res}}\; \frac{1}{w} (1+w)^{-1/2}
[z^{n-1}] (1+z)^{n-2} \sum_{k=0}^n
{n\choose k} \frac{1}{w^k} (1+z)^k
\\ = -\frac{2}{n} \times
\;\underset{w}{\mathrm{res}}\; \frac{1}{w} (1+w)^{-1/2}
[z^{n-1}] (1+z)^{n-2} (1+(1+z)/w)^n
\\ = -\frac{2}{n} \times
\;\underset{w}{\mathrm{res}}\; \frac{1}{w^{n+1}} (1+w)^{-1/2}
[z^{n-1}] (1+z)^{n-2} (1+w+z)^n
\\ = -\frac{2}{n} \times
\;\underset{w}{\mathrm{res}}\; \frac{1}{w^{n+1}} (1+w)^{-1/2}
[z^{n-1}] (1+z)^{n-2}
\sum_{q=0}^n {n\choose q} (1+w)^q z^{n-q}
\\ = -\frac{2}{n} \times
\sum_{q=1}^n {n\choose q} {q-1/2\choose n}
{n-2\choose q-1}.$$
Now observe that with $q\lt n$ (third binomial coefficient is zero
when $q=n$)
$${q-1/2\choose n} = \frac{1}{n!} (q-1/2)^\underline{n}
= \frac{1}{n!} \prod_{p=0}^{q-1} (q-1/2-p)
\prod_{p=q}^{n-1} (q-1/2-p)
\\ = \frac{1}{n! \times 2^n} \prod_{p=0}^{q-1} (2q-1-2p)
\prod_{p=q}^{n-1} (2q-1-2p)
\\ = \frac{1}{n! \times 2^n}
\frac{(2q-1)!}{(q-1)! \times 2^{q-1}}
\prod_{p=0}^{n-1-q} (-1-2p)
\\ = \frac{(-1)^{n-q}}{n! \times 2^n}
\frac{(2q-1)!}{(q-1)! \times 2^{q-1}}
\frac{(2n-1-2q)!}{(n-1-q)! \times 2^{n-1-q}}
\\= \frac{(-1)^{n-q}}{2^{2n-2}}
{n\choose q}^{-1} {2q-1\choose q-1} {2n-1-2q\choose n-q}.$$
We get for our sum
$$-\frac{1}{n \times 2^{2n-3}} \times
\sum_{q=1}^{n-1} (-1)^{n-q}
{2q-1\choose q-1} {2n-1-2q\choose n-q}
{n-2\choose q-1}
\\ = \frac{1}{n \times 2^{2n-3}} \times
\sum_{q=0}^{n-2} {n-2\choose q} (-1)^{n-2-q}
{2q+1\choose q} {2n-3-2q\choose n-q-1}.$$
This becomes
$$\frac{1}{n \times 2^{2n-3}} \times [z^{n-1}] (1+z)^{2n-3}
\sum_{q=0}^{n-2} {n-2\choose q} (-1)^{n-2-q}
{2q+1\choose q} z^q (1+z)^{-2q}
\\ = \frac{1}{n \times 2^{2n-3}}
\;\underset{w}{\mathrm{res}}\; \frac{1+w}{w} [z^{n-1}] (1+z)^{2n-3}
\\ \times
\sum_{q=0}^{n-2} {n-2\choose q} (-1)^{n-2-q}
\frac{1}{w^{q}} (1+w)^{2q} z^q (1+z)^{-2q}
\\ = \frac{1}{n \times 2^{2n-3}}
\;\underset{w}{\mathrm{res}}\; \frac{1+w}{w} [z^{n-1}] (1+z)^{2n-3}
\left(\frac{z(1+w)^2}{w(1+z)^2}-1\right)^{n-2}
\\ = \frac{1}{n \times 2^{2n-3}}
\;\underset{w}{\mathrm{res}}\; \frac{1+w}{w^{n-1}} [z^{n-1}] (1+z)
\left(z(1+w)^2-w(1+z)^2\right)^{n-2}
\\ = \frac{1}{n \times 2^{2n-3}}
\;\underset{w}{\mathrm{res}}\; \frac{1+w}{w^{n-1}} [z^{n-1}] (1+z)
(z-w)^{n-2} (1-wz)^{n-2}.$$
The first piece in $z$ is
$$[z^{n-1}] (z-w)^{n-2} (1-wz)^{n-2}
\\ = \sum_{q=1}^{n-2} {n-2\choose q} (-1)^{n-2-q} w^{n-2-q}
{n-2\choose n-1-q} (-1)^{n-1-q} w^{n-1-q}
\\ = - \sum_{q=1}^{n-2} {n-2\choose q} {n-2\choose q-1}
w^{2n-3-2q}.$$
Here we require
$$([w^{n-2}] + [w^{n-3}]) w^{2n-3-2q}$$
We get $q=(n-1)/2$ in the first case and $q=n/2$ in the second. As
this is a pair of an integer and a fraction clearly only one of these
extractors can return a non-zero value.
The second piece in $z$ is
$$[z^{n-2}] (z-w)^{n-2} (1-wz)^{n-2}
\\ = \sum_{q=0}^{n-2} {n-2\choose q} (-1)^{n-2-q} w^{n-2-q}
{n-2\choose n-2-q} (-1)^{n-2-q} w^{n-2-q}
\\ = \sum_{q=0}^{n-2} {n-2\choose q} {n-2\choose q}
w^{2n-4-2q}.$$
Solving for $q$ again we require
$$([w^{n-2}] + [w^{n-3}]) w^{2n-4-2q}$$
getting $q=n/2-1$ and $q=(n-1)/2.$
Supposing that $n$ is odd i.e. $n=2m+1$ we thus have
$$-{2m-1\choose m} {2m-1\choose m-1} +
{2m-1\choose m} {2m-1\choose m} = 0,$$
and we have proved the second part of the claim.
On the other hand with $n=2m$ even we collect
$$-{2m-2\choose m} {2m-2\choose m-1}
+ {2m-2\choose m-1} {2m-2\choose m-1}
\\ = {2m-2\choose m-1}^2 \left(1 - \frac{m-1}{m}\right)
= \frac{m^2} {(2m-1)^2} {2m-1\choose m}^2
\frac{1}{m}
\\ = \frac{m^2} {(2m-1)^2} \frac{m^2}{(2m)^2} {2m\choose m}^2
\frac{1}{m}
= \frac{1}{4} \frac{m} {(2m-1)^2} {2m\choose m}^2.$$
Restoring the factor in front we obtain
$$\frac{1}{n \times 2^{2n-3}}
\frac{1}{4} \frac{m} {(2m-1)^2} {2m\choose m}^2
= \frac{1}{2^{2n}} \frac{1} {(2m-1)^2} {2m\choose m}^2
\\ = \frac{1}{2^{4m}} \frac{1} {(1-2m)^2} {2m\choose m}^2$$
This is
$$\bbox[5px,border:2px solid #00A000]{
\left[\left(\frac{1}{4}\right)^m
{2m\choose m} \frac{1}{1-2m}\right]^2}$$
as was to be shown.
The confluent hypergeometric functions are related to the generalized Laguerre polynomials:
\begin{align}
F(-n;x;t)&=\frac{\Gamma(n+1)\Gamma(x)}{\Gamma(x+n)}L_n^{(x-1)}(t)
\end{align}
so
\begin{equation}
H_{n,m}(x)=\frac{n!m!\Gamma^2(x)}{\Gamma(x+n)\Gamma(x+m)}\int_0^\infty t^{x-1}e^{-t}\ln t L_n^{(x-1)}(t)L_m^{(x-1)}(t)\,dt
\end{equation}
The orthogonality relation for the Laguerre polynomials reads
\begin{equation}
\int_0^\infty t^{x-1}e^{-t}L_n^{(x-1)}(t)L_m^{(x-1)}(t)\,dt=\frac{\Gamma(n+x)}{n!}\delta_{n,m}
\end{equation}
It can be differentiated with respect to $x$ to obtain
\begin{equation}
\int_0^\infty t^{x-1}e^{-t}\ln t L_n^{(x-1)}(t)L_m^{(x-1)}(t)\,dt+\int_0^\infty t^{x-1}e^{-t}\frac{d}{dx}\left[L_n^{(x-1)}(t)L_m^{(x-1)}(t)\right]\,dt=\frac{\Psi(n+x)\Gamma(n+x)}{n!}\delta_{n,m}
\end{equation}
From the differentiation relation
\begin{equation}
\frac{d}{dx}L_n^{(x-1)}(t)=\sum_{k=0}^{n-1}\frac{L_k^{(x-1)}(t)}{n-k}
\end{equation}
and recognizing the definition of $H_{n,m}(x)$, we have thus
\begin{align}
\frac{\Gamma(n+x)\Gamma(m+x)}{n!m!\Gamma^2(x)}H_{n,m}(x)&+\sum_{k=0}^{n-1}\frac{1}{n-k}\int_0^\infty t^{x-1}e^{-t}L_k^{(x-1)}(t)L_m^{(x-1)}(t)\,dt\\
&+\sum_{k=0}^{m-1}\frac{1}{m-k}\int_0^\infty t^{x-1}e^{-t}L_k^{(x-1)}(t)L_n^{(x-1)}(t)\,dt\\
&=\frac{\Psi(n+x)\Gamma(n+x)}{n!}\delta_{n,m}
\end{align}
using the orthogonality relation and supposing that $n> m$, only one term in the sums survives, while there is no if $n=m$:
\begin{equation}
\frac{\Gamma(n+x)\Gamma(m+x)}{n!m!\Gamma^2(x)}H_{n,m}+\frac{1}{n-m}\frac{\Gamma(m+x)}{m!} \left( 1-\delta_{n,m} \right)=\frac{\Psi(n+x)\Gamma(n+x)}{n!}\delta_{n,m}
\end{equation} which is the proposed expression.
Best Answer
As you said in the comment, you are only trying to replace $\nu$ and $\mu$ (but not $L$) by complex numbers. You do not need the $\Gamma$-function for that; a binomial coefficient of the form $\dbinom{x}{k}$ is defined in the usual way (namely, as $\dfrac{x\left(x-1\right)\cdots\left(x-k+1\right)}{k!}$) whenever $x$ is a complex number and $k$ is a nonnegative integer. (The $\Gamma$-function only becomes necessary if you want to extend it nontrivially to non-integer values of $k$; even then, it is not clear whether such an extension is the most useful one.)
Furthermore, your identity (1) is an equality between two polynomials in $\nu$ and $\mu$ (when $L$ is held constant). The "polynomial identity trick" says that if such an equality holds whenever $\nu$ and $\mu$ are nonnegative integers, then it must also hold for arbitrary complex numbers $\nu$ and $\mu$. Since you (presumably) have a proof of (1) in the case whenever $\nu$ and $\mu$ are nonnegative integers, you thus automatically obtain a proof of (1) for arbitrary complex numbers $\nu$ and $\mu$. No further argument is required (though, of course, you can choose to come up with a different proof).