@Wadim Zudilin, could you tell me why you apply hypergeometric functions transforms? I think your idea may lead us to the solution.

@Wilberd van der Kallen, I have verified via maple the identity $\sum_{i=0}^m\sum_{j=0}^m\frac{3 (-1)^{k+j}{2k \choose k }{j \choose i}{ m \choose i }{i \choose m-j }}{2(2i-1)(2j+1)(2m-2i-1)}=0$ for $m\ge 2$, which is interesting.

@ Peter Mueller, Comparing the leading coefficient, we have $\sum_{i=0}^{m}{2m\choose m}{m\choose i}^2\frac{3}{(2i-1)(2m-2i-1)(2m+1)}$ is an integer. I think this sum has not a closed form. However, it is not difficulty to prove that it is an integer.

@Martin Rubey, The recurrence you gave was true only if $x=1$. For $x\ge 2$, $P_m(x)$ no longer satisfy the recurrence.