Fix $m,n \in \mathbb{Z}\setminus \{0\}$ and consider the subspace $$ A = \{(z_1,z_2) \in \mathbb{C}^2 :|z_1|=|z_2| = 1, z_1^m = z_2^n\} $$ of the $2$-torus. Is this space homeomorphic to a circle?
Thinking of $S^1$ as the unit circle in $\mathbb{C}$ there's a map $S^1 \to A$ given by $z \mapsto (z^n,z^m)$. Is this map obviously bijective?
Fix integers $m\neq 0$ and $n$. Our goal is to find all solutions to the equation $z_1^n = z_2^m$ with $(z_1,z_2)\in T^2$.
As you noted, the curve $S^1\ni z\mapsto (z_1, z_2) = (z^m,z^n)$ parameterizes infinitely many solutions to the equation $z_1^n = z_2^m$. One can potentially make more solutions via the following procedure: if $\alpha$ is an $n$-th root of unity and $\beta$ is an $m$-th root of unity, then any point of the form $$(z_1,z_2) = (\alpha z^m, \beta z^n) \hspace{1 in} (\ast)$$ is also a solution. We will prove that points of this form comprise all solutions, and we'll figure out all redundancies.
Here is the main result:
Let's prove it.
Proposition 1: All solutions have the form $(\ast)$. One can even take $\alpha=1$.
Proof: Suppose $(z_1,z_2)$ solves $z_1^n = z_2^m$. Because $m\neq 0$, there is an $|m|$-th root of $z_1$. Choose $z$ to be any such $|m|$-th root of $z_1$. Then $$(z^n)^m = (z^m)^n = z_1^n = z_2^m.$$ Thus, $z_2\overline{z}^n = \beta$ is an $m$-th root of unity. Then $$(z^m, \beta z^n) = (z_1, z_2)$$ as claimed. $\square$
So, we have a way of describing all solutions: every solution has the form $(z_1,z_2) = (z^m, \beta z^n)$ for some $z\in S^1$ and $\beta$ an $m$-th root of unity.
But there can be some redundancy in this description.
Proposition 2: Suppose $\beta_1,\beta_2$ are $|m|$-th roots of unity. If $\beta_1 = \beta^n \beta_2$ for some $|m|$-th root of unity $\beta$, then the two curves $z\mapsto (z^m, \beta_1 z^n)$ and $w \mapsto (w^m, \beta_2 w^n)$ have precisely the same image.
Conversely, if these two curves intersect in any point, then $\beta_1 = \beta^n \beta_2$ for some $|m|$-th root of unity $\beta$.
Proof: Suppose first that $\beta_1 = \beta^n \beta_2$. Now, given $w$, set $z = \beta^{-1} w$. Then $$(z^m, \beta_1 z^n) = ((\beta^{-1} w)^m, \beta_1 (\beta^{-1} w)^n) = (w^m, \beta_1 \beta^{-n} w^n) = (w^m, \beta_2 w^n),$$ so every point on the $w$-paramaterized curve is on the $z$-paramaterized curve. Given $z$, set $w = \beta z$ to get the reverse inclusion. This completes the proof of the first statement.
To prove the second, let's assume the two curves intersect. So we have equality $(z^m, \beta_1 z^n) = (w^m, \beta_2 w^n)$. From the equation $z^m = w^m$, we conclude that $z = \beta w$ where $\beta$ is an $|m|$-th root of unity. Substituting this into the second equation $\beta_1 z^n = \beta_2w^n$, we find that $\beta_1 (\beta)^n w^n = \beta_2 w^n$, so that $\beta_1 (\beta)^n = \beta_2$. $\square$
Here is one way to interpret this. The set of $|m|$-th roots of unity is isomorphic (as a group) to $\mathbb{Z}_m$. We have a subgroup $n\mathbb{Z}_m$ (which is automatically normal) given as all elements of the form $n\cdot x$ with $x\in \mathbb{Z}_m$. Then Proposition $2$ says the solution to the equation $z_1^n = z_2^m$ consists of disjoint closed curves, where the connected components of the solution set are in bijection with $\mathbb{Z}_m/H_n$. Thus, to count connected components, we need to determine the order of $\mathbb{Z}_m/H_n$.
Of course, to do this, we need only determine the order of $H_n$. To that end, note that $H_n$ is cyclic generated by $n$ (since $H_n$ is the homomorphic image of $\mathbb{Z}_m$ under the "times $n$" map.) The order of $H_n$ is therefore the order of $n\in \mathbb{Z}_m$. From, e.g., this MSE question, the order is $\frac{m}{\gcd(m,n)}$. Thus, the order of $\mathbb{Z}_m/H_n$ is $\frac{m}{\frac{m}{\gcd(m,n)}} = \gcd(m,n)$.