Croatian Mathematical Olympiad

Note that $2^m\equiv 1(\mathrm{mod}7)$, 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 $B\equiv 1(\mathrm{mod}4)$, i.e. $7b^2\equiv 1(\mathrm{mod}4)$, 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. $2^k\equiv 1(\mathrm{mod}3)$, 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: $$\begin{array}{rl}{2}^l-1& =c^2\\ 2^l+1& =3d^2\end{array}\text{or}\begin{array}{rl}{2}^l-1& =3c^2\\ 2^l+1& =d^2,\end{array}$$ where $c$ and $d$ are relatively prime positive integers.

In the first option, if $l\ge 2$ we have $3d^2\equiv 1(\mathrm{mod}4)$, 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 $(m,n)=(6,3)$.