= Solution
There is a normalization conflict in the source: the printed <Rodrigues' formula> gives the standard <Generalized Laguerre polynomial> $P_n=L_n^{(m)}$ with $P_n(0)=\binom{n+m}{n}$, which is generally not one. The norm requested here corresponds to that Rodrigues normalization. A solution normalized to one at zero is $\widehat P_n=P_n/\binom{n+m}{n}$.
Set $w=x^me^{-x}$ and $p=x^{m+1}e^{-x}$. The equation becomes $(pP_n')'+nwP_n=0$. Multiply the equations for $P_n,P_k$ by the opposite <polynomial>, subtract and integrate. The boundary term $p(P_n'P_k-P_k'P_n)$ vanishes at zero and infinity, so
$$
(n-k)\int_0^\infty wP_nP_k\,dx=0.
$$
This proves <orthogonality> for $k\ne n$ under either normalization.
For the squared norm, use the printed <Rodrigues' formula> and integrate by parts $n$ times:
$$
\int_0^\infty wP_n^2\,dx
=\frac1{n!}\int_0^\infty P_nD^n(x^{n+m}e^{-x})\,dx
=\frac{(-1)^n}{n!}\int_0^\infty P_n^{(n)}x^{n+m}e^{-x}\,dx.
$$
All boundary terms vanish: the required derivatives have a positive power of $x$ at zero and exponential decay at infinity. The leading coefficient is $(-1)^n/n!$, so $P_n^{(n)}=(-1)^n$. The <Gamma function> integral gives
$$
\boxed{\int_0^\infty x^me^{-x}P_n(x)^2\,dx=\frac{(n+m)!}{n!}.}
$$
For the source's alternative unit-at-zero normalization the correct answer instead is
$$
\boxed{\int_0^\infty x^me^{-x}\widehat P_n(x)^2\,dx=\frac{(n+m)!}{n!\binom{n+m}{n}^2}.}
$$
Both cannot equal the printed norm when $m,n>0$. This distinction is the <normalization of generalized Laguerre polynomials>.
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