Let $y_1<z_1<y_2<z_2<\dots<y_l<z_l<y_{l+1}$ be roots of those two polynomials ($y$'s of $f=P_{l+1}^{m}$, $z$'s of $g=P_{l}^m$). I hope that they are real and alternate as written (we should think why or search it in the literature.)

Then for Wronskian we have $f'g-g'f=0$ iff $f'/f-g'/g=0$, i.e. $$ \sum \frac1{x-y_i}=\sum \frac1{x-z_i}. $$ But it is impossible for real $x$. If, say, $y_k<x<z_k$ for some $k$, then $1/(x-y_i)>1/(x-z_{i-1})$ for $i=2,\dots,{l+1}$ and $1/(x-y_1)>0$, thus LHS exceed RHS.

Let $y_1< z_1< y_2< z_2< \dots< y_{l-m}\leq z_{l-m}< y_{l-m+1}$ be roots of polynomials $f=(1-x^2)^{-m/2} P_l^m$ and $g=(1-x^2)^{-m/2} P_{l+1}^m$ ($y$'s are roots of $g$, $z$'s are roots of $f$). They are real, belong to $(-1,1)$ and alternate as written because $f$, $g$ are orthogonal polynomial in weighted $L^2_w(-1,1)$ with weight $w=(1-x^2)^m$.

Wronskian vanishes (in a point different from $\pm 1$) iff $(P_{l+1}^m/P_l^m)'=0$, i.e. iff $(g/f)'=0$, i.e. iff $f'/f=g'/g$ i.e. iff $$ \sum \frac1{x-y_i}=\sum \frac1{x-z_i}. $$ But it is impossible for real $x$. If, say, $y_k<x<z_k$ for some $k$, then $1/(x-y_i)>1/(x-z_{i-1})$ for $i=2,\dots,{l-m+1}$ and $1/(x-y_1)>0$, thus LHS exceed RHS.