I won't delete my earlier answer, but here is a much simpler proof that your groups have trivial Schur Multiplier. This follows from the fact that they have balanced $2$-generator $2$-relator presentations $$\langle x,y \mid yxy^{-1}=x^2,y^n=1 \rangle.$$
Note that the two relations in this presentation imply $x = y^nxy^{-n} = x^{2^n}$, so $x^{2^n-1} = 1$, and hence the presentation defines your semidirect product $C_{2^n-1} \rtimes C_n$.
Note that in the general theory of Schur multipliers and covering groups, $G$ is not restricted to being a finite group.
For a covering group $H$ of $G$, choose elements $h_i \in H$ (for $i$ in some index set) such that their images in $H/H'$ generate $H/H'$. Since $H/Z(H)$ is the fixed group $G$, we can choose the $h_i$ to map onto specific elements of $G$ (i.e. independently of the choice of covering group $H$). Then, as I think you know, then values of the commutators $[h_i,h_j] \in H'$ do not depend on the choice of $H$. (In other words, the different covering groups of $G$ are all isoclinic.)
So, if $R$ is a set of relators of $H/H'$ that define $H/H'$ as an abelian group, then the isomorphism class of $H$ is determined by the values of $r_j \in R$ in $H'$, and these differ in different covering groups only by elements of $Z$.
In other words, if $r_j = g_j$ in $H$, and $H_1$ is another cover of of $G$, then $r_j = g_jz_j$ in $H_1$, for some $z_j \in Z$.
(Of course, if $H/H'$ is finitely generated, then we can choose the $h_i$ such that their images generate the cyclic direct factors of $H/H'$, and then we can choose powers of $h_i$ as the relators of $H/H'$ as abelian group.)
Consider the abelian group extension $E$ of $Z$ by $H/H'$ defined by the relations $r_j = z$. This extension is split if and only if there exist $z_j' \in Z$ such that, when we replace the generators of $h_i$ of $H$ by $h_iz_j'$, the relations $r_j$ in $E$ become trivial in $E$, but since the $z_j$ are central in $H$, this is equivalent to the relations $r_j = g_jz_j$ in $H_1$ changing to $r_j = g_j$, in which case $H$ and $H_1$ are isomorphic.
So equal elements of ${\rm Ext}(H/H',Z)$ (the equivalence classes of abelian group extensions of $Z$ by $H/H'$) induce isomorphic covering groups, and hence the number of isomorphism classes of covering groups is at most $|{\rm Ext}(H/H',Z)|$ which, in your notation in the finite case, is equal to $\prod_{i,j} \gcd (n_i,m_j)$.
Of course this is an inequality, and it is possible for Schur covers to be isomorphic even when the corresponding elements of ${\rm Ext}(H/H',Z)$ are distinct. I believe that such isomorphisms are induced by automorphisms of $G$. Interesting examples are the symmetric groups $S_n$. We have $|M(S_n)| = 2$ for all $n \ge 4$, and, for $n \ne 6$, there are two isomorphism classes of covering groups, but for $n=6$ there is only one, with the exceptional outer automorphism of $S_6$ inducing an isomorphism between them.
Best Answer
Here is the complete list of Schur trivial finite simple groups :
Nearly all the finite simple groups of Lie type for characteristic 2 are Schur trivial
$ A_n(2^m) $ i.e. $ PSL(n+1,2^m) $ Schur trivial exactly when $ 2^m-1 $ is coprime to $ n+1 $, except cases $ (n=1,m=1),(n=1,m=2),(n=2,m=1),(n=2,m=2),(n=3,m=1) $ which are not Schur trivial
$ B_n(2^m) \cong C_n(2^m) $ for $ n \geq 2 $ and all $ m $, except the cases $ n=2,m=1 $ and $ n=3, m=1 $.
$ D_n(2^m) $ for $ n \geq 4 $ and all $ m $, except the case $ (n=4,k=1) $
$ G_2(2^m) $ for $ m \geq 3 $
$ F_4(2^m) $ for $ m \geq 2 $
$ E_6(2^m) $ Schur trivial exactly when $ 2^m-1 $ is coprime to $ 3 $
$ E_7(2^m) $ all $ m $
$ E_8(2^m) $ all $ m $
$ ^2A_n(2^m) $ i.e. $ PSU(n+1,2^m) $ for $ n \geq 2 $, Schur trivial exactly when $ 2^m+1 $ is coprime to $ n+1 $ except cases $ (n=3,m=1),(n=5,m=1) $
$ ^2D_n(2^m) $ for $ n \geq 4 $ and all $ m $
$ ^2E_6(2^m) $ , Schur trivial exactly when $ 2^m+1 $ is coprime to $ 3 $
$ ^3D_4(2^m) $ for all $ m $
$ ^2B_2(2^{2n+1})$ for $ n \geq 2 $
$ ^2F_4(2^{2n+1})$ for $ n \geq 1 $, also Tits group $ ^2F_4(2)' $
For odd characteristic, some finite simple groups of Lie type are still Schur trivial, but it's a lot less. They are: (here $ q $ is an odd prime power)
$ A_n(q) $ i.e. $ PSL(n+1,q) $ Schur trivial exactly when $ q-1 $ is coprime to $ n+1 $
$ G_2(q) $ for $ q \neq 3 $
$ F_4(q) $
$ E_6(q) $
$ E_8(q) $
$ ^2A_n(q) $ i.e. $ PSU(n+1,q) $, for $ n \geq 2 $, Schur trivial exactly when $ q+1 $ is coprime to $ n+1 $, except case $ (n=3,q=3) $
$ ^2E_6(q) $ , Schur trivial exactly when $ q+1 $ is coprime to $ 3 $
$ ^3D_4(q) $
Ree groups $ ^2G_2(3^{2n+1})$ for $ n \geq 1 $, also $ ^2G_2(3)'\cong PSL(2,8) $
That is all the Schur trivial finite simple groups of Lie type.
Beyond that, of the 26 exceptional groups exactly these 13 are Schur trivial
$ M_{11}, M_{23}, M_{24} $
$ J_1,J_4 $
$ Co_2,Co_3 $
$ Fi_{23} $
$ He $
$ HN $
$ Ly $
$ Th $
$ M $
Finally no simple alternating group is Schur trivial. But all the cyclic simple groups are Schur trivial