Note that $2^m \equiv 1 \pmod{7}$, which means that $m = 3k$ for some positive integer $k$. Now we have
$$2^{3k} - 1 = (2^k - 1)(2^{2k} + 2^k + 1) = 7n^2.$$
Denote $A = 2^k - 1$ and $B = 2^{2k} + 2^k + 1$. Let $d$ be the greatest common divisor of $A$ and $B$, and note that $d$ is odd. Furthermore, $d$ divides $B - A^2 = 2^k + 2^{k+1} = 3 \cdot 2^k$, and since $d$ is odd, it follows that $d = 1$ or $d = 3$.
Therefore, we have exactly four possibilities for factorising $A$ and $B$. In all cases we assume that $a$ and $b$ are relatively prime positive integers.

Case 1. $A = 7a^2$ and $B = b^2$. This is not possible because $(2^k)^2 < B < (2^k + 1)^2$, i.e. $B$ cannot be a perfect square.

Case 2. $A = a^2$ and $B = 7b^2$. If $k \ge 2$, note that $B \equiv 1 \pmod{4}$, i.e. $7b^2 \equiv 1 \pmod{4}$, which is not possible. Therefore, $k = 1$ and we get $(m, n) = (3, 1)$.

Case 3. $A = 21a^2$ and $B = 3b^2$. Analogously to the previous case we can conclude that $k=1$ necessarily, but that is not possible in this case, since $21a^2 = A = 2^1 - 1 = 1$.

Case 4. $A = 3a^2$ and $B = 21b^2$. Therefore, $A$ is divisible by $3$, i.e. $2^k \equiv 1 \pmod{3}$, from which it follows that $k=2l$ for some positive integer $l$. Now we have
$$(2^l - 1)(2^l + 1) = 3a^2,$$
and since $2^l - 1$ and $2^l + 1$ are relatively prime, we have only two possibilities:
$$\begin{aligned} 2^l - 1 &= c^2 \\ 2^l + 1 &= 3d^2 \end{aligned} \quad \text{or} \quad \begin{aligned} 2^l - 1 &= 3c^2 \\ 2^l + 1 &= d^2, \end{aligned}$$
where $c$ and $d$ are relatively prime positive integers.

In the first option, if $l \ge 2$ we have $3d^2 \equiv 1 \pmod{4}$, which is not possible. Therefore, $l=1$ and we get another solution $(m,n) = (6,3)$.

In the second option, we have $2^l = (d-1)(d+1)$. Since the greatest common divisor of $d-1$ and $d+1$ is at most $2$ and their product is a power of $2$, we can conclude that $d-1=2$, i.e. that $l=3$, but then $3c^2 = 2^3 - 1 = 7$, which is not possible.

Finally, all solutions of the given equation are $(m,n) = (3,1)$ and $(m,n) = (6,3)$.