Given the Cauchy-like matrix $$ \mathbf X_M(q) = \left[ \frac{2}{\pi} \frac{ \Gamma\!\left(m - \frac{1}{2} \right)\Gamma\!\left(n + \frac{1}{2} \right) }{ \Gamma(m)\,\Gamma(n) } \frac{m-\frac{3}{4}} {\left(m-\frac{3}{4}\right)^2-\left(n-\frac{1}{4}\right)^2} q^{m+n-1} \right]_{m,n=1}^{M}, $$ show that for $M \to \infty$ $$ \det[\mathbf I_{\infty} + \mathbf X_{\infty}^T(q) \, \mathbf X_{\infty}(q)] = (q^2)_{\infty}^{-1/4} = \prod_{k=1}^{\infty}\frac{1}{(1-q^{2k})^{1/4}},\qquad (1) $$
with the Euler Gamma function $\Gamma(x)$ and the $q$-Pochhammer symbol $(q)_{\infty}$, while $\mathbf X^T$ denotes the transpose of $\mathbf X$. Note that this conjecture appeared in chapter 5.1 of [1], with $q = \exp(-\pi \rho)$, and I still have no idea how to prove it.
Motivated by a related question 259104 (see comments), I looked at similar determinant expressions for $\mathbf X_\infty(q)$, and by comparing the series expansions in $q$ found $$ \det[\mathbf I_{\infty} \pm \mathbf X_{\infty}(q)] = \det[\mathbf I_{\infty} \mp \mathbf X_{\infty}(-q)] = \prod_{k=1}^{\infty}\frac{(1 \mp q^{2k-1})^{1/2}}{(1-q^{2k})^{1/8}}, \qquad(2) $$ such that $$ \det[\mathbf I_{\infty} - \mathbf X_{\infty}^2(q)] = \prod_{k=1}^{\infty}\frac{(1 - q^{4k-2})^{1/2}}{(1-q^{2k})^{1/4}} = \prod_{k=1}^{\infty}\frac{1}{(1-q^{2k})^{(-1)^k/4}},\qquad(3) $$ which is an alternating product version of (1). Note the apparent similarity to (1) and (2) in 259104, which are the $k=1$ terms of our products. So the question is: How can one prove these identities?
See also related MathOverflow questions 199922 and 236323 and 259104.
[1] Hucht, Alfred, The square lattice Ising model on the rectangle. II: Finite-size scaling limit, J. Phys. A, Math. Theor. 50, No. 26, Article ID 265205, 23 p. (2017). ZBL1369.82005.
