We want to prove the formula
$$\frac{\int_{-\infty }^{\infty } \frac{e^{-x^2} H_n(x){}^2}{1+x^2} \, dx}{2^n n!}=\int_{-\infty }^{\infty } \frac{\left(\frac{1-x^2}{1+ x^2}\right)^n e^{-x^2}}{1+x^2} \, dx$$ where $H_n(x)$ is the Hermite polynomial. The usual procedure of invoking the identity $$\frac{\exp \left(\frac{2 x^2 t}{1+t}\right)}{\sqrt{1-t^2 }}=\sum _{n=0}^{\infty } \frac{t^n H_n(x){}^2}{2^n n!}$$ Leads to the relation $$\frac{\int_{-\infty }^{\infty } \frac{e^{-x^2} \exp \left(\frac{2 t x^2}{1+t}\right)}{1+x^2} \, dx}{\sqrt{1-t^2}}=\sum _{n=0}^{\infty } \frac{t^n \int_{-\infty }^{\infty } \frac{e^{-x^2} H_n(x){}^2}{1+x^2} \, dx}{2^n n!}$$ It is not clear how to match powers of $t$ on both sides and obtain the desired formula.
Hint: Upon invoking your summation identity, your new integrand contains the exponential factor $$\exp(-x^2)\exp\left(\frac{2tx^2}{1+t}\right)=\exp\left(-\frac{1-t}{1+t} x^2\right).$$
The coefficient $\dfrac{1-t}{1+t}$ may be eliminated by a judicious substitution. It remains to organize the new integrand so as to permit expanding in powers of $t$, and this is straightforward.