I used the Rodrigues formula and integrated by parts: $$ \dfrac{1}{2^{l}l!}\int_{0}^{1}x^{m}\dfrac{d^{l}}{dx^{l}}(x^2 - 1)^{l}dx = \dfrac{1}{2^{l}l!}[ x^{m}\dfrac{d^{l-1}}{dx^{l-1}} (x^{2} - 1)^{l} - \int_{0}^{1} m x^{m-1}\dfrac{d^{l-1}}{dx^{l-1}} (x^{2} - 1)^{l}dx] $$ The first term vanishes due to the integration extremes. Then integrating by parts m times: $$\dfrac{1}{2^{l}l!}[(-1)^{m}m!\int_{0}^{1}\dfrac{d^{l-m}}{dx^{l-m}}(x^{2} - 1)^{l}dx]$$
But I don't know how to develop from here!
Hint:
It is clear from the RHS of your equation that the inequality $m\ge l-1$ is assumed. To incorporate the case $m=l-1$ we will follow the same path as you proposed but stop after $l-1$ integrations by parts to obtain:
$$\dfrac{1}{2^{l}l!}\left[\frac{m!(-1)^{l-1}}{(m-l+1)!}\int_{0}^{1}x^{m-l+1}\frac d{dx}(x^{2}-1)^{l}dx\right]=\dfrac{1}{2^{l}l!}\left[\frac{m!\,2l}{(m-l+1)!}\int_{0}^{1}x^{m-l+2}(1-x^{2})^{l-1}dx\right].$$
It remains to prove: $$ \int_{0}^{1}x^{m-l+2}(1-x^{2})^{l-1}dx=\frac{(m-l+1)!!\,(2l-2)!!}{(m+l+1)!!}. $$
Can you take it from here?