We want $F(z,w)$ a generating function such that
$$F(z,w) = \sum_{x=0}^{\infty} \sum_{y=0}^{\infty} (y^2-1)^x z^xw^y \tag{1}$$
$(y^2-1)^x z^x$ is a power series.
$$ F(z,w) = \sum_{y=0}^{\infty} \frac{w^y}{1 - (y^2-1)z} \tag{2}$$
$$ F(z,w) = -\frac1{z} \sum_{y=0}^{\infty} \frac{w^y}{y^2 - 1 - \frac1{z}} \tag{3}$$
$$ F(z,w) = -\frac1{z} \sum_{y=0}^{\infty} \frac{w^y}{\left(y - \sqrt{1 + \frac1{z}}\right)\left(y + \sqrt{1 + \frac1{z}}\right)} \tag{4}$$
$$ F(z,w) = \frac1{2z\sqrt{1 + \frac1{z}}} \left[ \sum_{y=0}^{\infty} \frac{w^y}{y + \sqrt{1 + \frac1{z}}} - \sum_{y=0}^{\infty} \frac{w^y}{y - \sqrt{1 + \frac1{z}}} \right] \tag{5}$$
Now we need a generating function such that
$$G(w) = \sum_{y=0}^{\infty} \frac{w^y}{y + c} \tag{6}$$
$$\sum_{y=0}^{\infty} w^y = \frac1{1-w} \tag{7}$$
$$w^{c-1}\sum_{y=0}^{\infty} w^y = \sum_{y=0}^{\infty} w^{y+c-1} = \frac{w^{c-1}}{1-w} \tag{8}$$
$$\int\sum_{y=0}^{\infty} w^{y+c-1} dw = \sum_{y=0}^{\infty} \frac{w^{y+c}}{y+c} = \int \frac{w^{c-1}}{1-w} dw \tag{9}$$
$$w^{-c}\sum_{y=0}^{\infty} \frac{w^{y+c}}{y+c} = \sum_{y=0}^{\infty} \frac{w^{y}}{y+c} = w^{-c}\int \frac{w^{c-1}}{1-w} dw \tag{10}$$
From wolfram +c and wolfram -c
$$\int \frac{w^{c-1}}{1-w} dw = \frac{w^c (c w \, _{2}F_{1}(1, c+1, c+2, w)+c+1)}{c (c+1)}+constant \tag{11}$$
Where $_{2}F_{1}$ is a hypergeometric function.
$$ F(z,w) = \frac1{2\,z\,c} \left[ \frac{c w \, _{2}F_{1}(1, c+1, c+2, w)+c+1}{c (c+1)} + \frac{c w \, _{2}F_{1}(1, -c+1, -c+2, w)+c-1}{c (c-1)}\right] \tag{12}$$
Where $c = \sqrt{1 + \frac1{z}}$
Sanity check wolfram
