The correct condition is that $m$ and $n$ are coprime, that is, have no common factors. In particular, this means that $C_2 \times C_3 \cong C_6$. (Sometimes it is hard to compare two multiplication tables by inspection, though it can help identifying two isomorphic groups from their tables by reordering the elements.)
Hint Given two groups $G, H$ and elements $a \in G, b \in H$, the order of element $(a, b)$ is the smallest number, $\text{lcm}(k, l)$ divisible by both the order of $l$ of $a$ and the order $k$ of $b$.
In particular, suppose $G$ and $H$ are cyclic, say $G = C_n$, $H = C_m$. Then, if $a$ and $b$ are generators of $C_n$, $C_m$, respectively, by definition they have respective orders $n, m$ and hence $(a, b) \in C_n \times C_m$ has order $\text{lcm}(m, n)$.
So, if $m, n$ are coprime, this order is $\text{lcm}(m, n) = mn$.
Conversely, if $m, n$ are not coprime, then since any $a \in C_n$ has order $l$ dividing $n$ and any $b \in C_m$ has order $k$ dividing $m$, we have that the generic element $(a, b) \in C_n \times C_m$ has order $\text{lcm}(k, l) \leq \text{lcm}(m, n) < mn$. In particular, no element has order $mn$, so $C_m \times C_n$ is not cyclic.
For example, the elements $1 \in C_2$ and $1 \in C_3$ generate their respective groups, and the powers of $(1, 1) \in C_2 \times C_3$ are
$$(1, 1), (0, 2), (1, 0), (0, 1), (1, 2), (0, 0),$$
so $(1, 1)$ generates all of $C_2 \times C_3$, which is hence isomorphic to $C_6$.
I just discovered that Magma supports "GroupNames", so I tried the following instruction
G:=Group("C2^5:C7:He3");
and then I checked some properties:
Order(G);
Center(G);
Centraliser(G,Sylow(G,7));
Centraliser(G,Sylow(G,2));
T:=Sylow(G,3);
Centraliser(G,T.1);
Centraliser(G,T.2);
Centraliser(G,T.1*T.2);
...and it seems that this is indeed my group.
Best Answer
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 $C_{2^n-1} \rtimes C_n$.