Let $A_n(x,\lambda)$ be the $n\times n$ matrix
$$\left[\binom{x}{2j-i+\lambda}\right]_{i,j=1}^n.$$
Let's "generalize to trivialize". Sometimes, generalizations offer more elbow room to maneuver, such as in the present problem.

**Lemma.** We have the determinantal formula
$$\det A_n(x,\lambda)=2^{\binom{n}2}\prod_{i=1}^n\binom{n+x-1}{\lambda+2i-1}\binom{n+x-1}{n-i}^{-1}.$$
**Proof.** We employ the Method of Condensation. Denote $k\times k$ sub-matrices with left-corner at $(a,b)$:
$$Z_k^{a,b}(x,\lambda)=\left[\binom{x}{2(j+b)-(i+a)+\lambda}\right]_{i,j=1}^k.$$
Notice that $Z_n^{0,0}(x,\lambda)=A_n(x,\lambda)$. In general $Z_k^{a,b}(x,\lambda)=A_k(x,\lambda+2b-a)$, therefore the inductive proofs neatly work with this Dodgson's recursive relation
$$\det Z_n^{a,b}=\frac{\det Z_{n-1}^{a,b}\cdot\det Z_{n-1}^{a+1,b+1}-\det Z_{n-1}^{a,b+1}\cdot\det Z_{n-1}^{a+1,b}}{\det Z_{n-2}^{a+1,b+1}}.$$
So, it remains to prove that the (explicit) formula on the RHS of the lemma *does* satisfy the same equation. However, this is quite a routine simplification (preferably with symbolic sofwares). The proof is complete. $\square$

**Corollary 1.** Your determinant evaluates as $2^{\binom{2m}2}$.

**Proof.** Take $n=2m-1, x=2m, \lambda=0$ in $A_n(x,\lambda)$ and simplify the RHS of the lemma. $\square$

Given a function $F(y_1,\cdots,y_{2m-1})$, denote the *constant term* of $F$ w.r.t. $y_i$ by $CT_i(F)$ and let $CT=\prod_{i=1}^{2m-1}CT_i$. We now register a nice consequence. A bonus, say to say.

**Corollary 2.** If $V(z_1,\dots,z_{2m-1})$ denotes the determinant of the Vandermonde matrix then
$$CT\left(\prod_{i=1}^{2m-1}y_i^{i-2}(1+y_i)^{2m}V(y_1^{-2},\dots,y_{2m-1}^{-2})\right)=2^{\binom{2m}2}.$$
**Proof.** Since $\binom{2m}{2j-i}=CT_i(y_i^{i-2j}(1+y_i)^{2m})$,
\begin{align} \det\left[\binom{2m}{2j-i}\right]
&=\det\left[CT_i(y_i^{i-2j}(1+y_i)^{2m})\right]
=CT\prod_{i=1}^{2m-1}y_i^i(1+y_i)^{2m}\det\left[y_i^{-2j}\right] \\
&=CT\prod_{i=1}^{2m-1}y_i^{i-2}(1+y_i)^{2m}\cdot\left(\det\left[(y_i^{-2})^{j-1}\right]\right) \\
&=CT\prod_{i=1}^{2m-1}y_i^{i-2}(1+y_i)^{2m}\cdot V(y_1^{-2},\dots,y_{2m-1}^{-2}).
\end{align}
The assertion follows from Corollary 1. $\square$

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