finding generating function of orthogonal polynomials through their moments

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I was studying a method to find the generating function of Orthogonal Polynomials through its moments. Please refer to the paper Use of Hermite's method to obtain generating functions for classical orthogonal polynomials [N.K. Thakare and H.C. Madhekar, Indian J. pure appl. Math., 13, 183-189 (1982)]

For example, lets $g(x,t)$ be the generating function of the Hermite polynomials, $H_n$, such that $$ g(x,t) = \sum_{n=0}^\infty \frac1{n!} H_n(x) t^n $$

The $k$-th moment of this equation is $$ I_k(t) = \int_{-\infty}^\infty dx\, e^{-x^2} x^k g(x,t) = \sum_{n=0}^k \frac{t^n}{n!} \int_{-\infty}^\infty dx\, e^{-x^2} x^k H_n(x) $$

The last part of this equality says that $I_k(t)$ must be a polynomial of degree $k$, because $H_n$ is orthogonal to every polynomial of degree $k<n$.

Under the change of variables $x=y+t$, we obtain $$ I_k(t) = \int_{-\infty}^\infty dy\, e^{-y^2} e^{-2yt-t^2}(y+t)^k g(x,t)$$

Here is my question: The paper of Thakare, cited above, claims that the only way $I_k$ is a polynomial is when $g(x,t) = e^{2yt+t^2} = e^{2xt-t^2}$. Why is this true?

One arrives to the following formula, $$ I_k(t) = \int_{-\infty}^\infty dy\, e^{-y^2} (y+t)^k $$ which actually gives a polynomial after explicit integration, but why is this procedure true for other orthogonal polynomials, e.g Laguerre or Lagrange?