- Method 1: recurrence relations and symmetries of the Hermite polynomials
Hint: I would use the relations
$$H_{n+1}(x)=xH_n(x)-H'_n(x), $$
$$H'_n(x)=nH_{n-1}(x) $$
and an induction argument, as follows. Let us suppose the steps $k=1,\dots,n-1$ are true; we want to show that
$$\int e^{-x^2}H_n^2(x)x^2dx=2^nn!\sqrt{\pi}(n+\frac{1}{2}). $$
Using the above relations the l.h.s is
$$\int e^{-x^2}H_n^2(x)x^2dx=\int e^{-x^2}x^2(xH_{n-1}(x)-H'_{n-1}(x))^2dx=
\int e^{-x^2}x^4H^2_{n-1}(x)dx \\
-2\int e^{-x^2}x^3H_{n-1}(x)H^{'}_{n-1}(x)dx
+\int e^{-x^2}x^2(H^{'}_{n-1}(x))^2dx=\\
\int e^{-x^2}x^4H^2_{n-1}(x)dx -2(n-1)\int e^{-x^2}x^3H_{n-1}(x)H_{n-2}(x)dx
+\\(n-1)^2\int e^{-x^2}x^2(H_{n-2}(x))^2dx. ~(*)
$$
Let us discuss $(*)$: the first 2 terms can be reduced using integration by parts on the products
$$e^{-x^2}x^4H^2_{n-1}(x)=-(-\frac{1}{2}(2x)e^{-x^2})x^3H_{n-1}(x),$$
$$e^{-x^2}x^3H_{n-1}(x)H_{n-2}(x)=-(-\frac{1}{2}(2x)e^{-x^2})x^2H_{n-1}(x)H_{n-2}(x), $$
while the last term can be evaluated by the induction hypothesis.
All we need is to remember that
$$\int e^{-x^2}x^{2q+1}H^2_{r}(x)dx=0,$$
$$\int e^{-x^2}x^{2q}H_{r}(x)H_{r-1}(x)dx=0,$$
for all $q\in\mathbb N$, $r\geq 1$ for symmetry. For the second equality we used
$H_r$ odd/even $\Leftrightarrow$ $H_{r-1}$ even/odd, which follows directly from the definition of the Hermite polynomials.
- Method 2: generating function
Using the generating function suggested by the OP, one needs to integrate w.r.t $x$ on the whole real axis both sides of
$$x^2e^{-x^2}e^{-\frac{t}{1+t}x^2}\frac{1}{\sqrt{1-t^2}}= \sum_{n=1}^\infty x^2e^{-x^2}\frac{H_n^2(x)}{2^nn!}t^n,$$
and expanding w.r.t. $t$. The integral on the l.h.s. is Gaussian and can be evaluated with standard techniques reducing to the case
$$\int_{-\infty}^\infty x^2e^{-\alpha(t)x^2}dx=\frac{1}{2}\sqrt{\frac{\pi}{\alpha(t)^3}}, ~(1)$$
for $\alpha(t)>0$ and $|t|<1$.
- Generating function with $\alpha(t)=\frac{1-t}{1+t}$
In the case under exam, multiplying both sides of the generating function identity w.r.t the product $x^2 e^{-x^2}$ ($e^{-x^2}$ is the weight for the integral identities involving the Hermite polynomials) and integrating, we arrive at (the exchange between summation and integration is allowed by the properties of Hermite poly.)
$$\frac{1}{\sqrt{1-t^2}} \int_{-\infty}^\infty x^2 e^{-x^2} e^{-\frac{t}{1+t}x^2}dx= \sum_{n=1}^\infty \int_{-\infty}^\infty x^2e^{-x^2}\frac{H_n^2(x)}{2^nn!}t^n; $$
to arrive at the thesis we write (using (1))
$$g(t):=\frac{1}{\sqrt{1-t^2}}\int_{-\infty}^\infty x^2e^{-\frac{1-t}{1+t}x^2}dx=\frac{1}{\sqrt{1-t^2}}\frac{1}{2}\sqrt{\frac{\pi}{\left(\frac{1-t}{1+t}\right)^3}}=\frac{\sqrt{\pi}}{2}\frac{1+t}{(1-t)^2}$$
for $|t|<1$ and we expand w.r.t $t$ at $t=0$ the function $g(t)$, obtaining
$$g(t)=\frac{\sqrt{\pi}}{2}\left(1+3t+\frac{1}{2!}10t^2+O(t^3)\right)$$
(the higher terms are easily computed, as well). This gives the thesis. I hope it helps.