Some discuss in here http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=409486 [Moderator edit: And here. ]

I have another competely different solution. Taking mod $m$, we get $m\mid 2\times 3^n$. Since if $(m,n)$ is a solution, $\left(\frac{2\times 3^n}{m},n\right)$ is also a solution, so WLOG assume $m=3^a$. Then we have $3^a+2\times 3^{n-a}=2^{n+1}-1$. If either $a<3$ or $n-a<3$, it is easy to check that $(a,n)=(2,3),(2,5)$ are the only solutions by simple bounding. So now assume both $a$ and $n-a$ are at least 3. Taking mod 27, we get $18\mid n+1$. If $a\leq n-a$, then $3^a+2\times 3^{n-a}=3^a(1+2\times 3^{n-2a})$. Taking mod 19, we get $19\mid RHS$, so $19\mid 1+2\times 3^{n-2a}\implies n-2a\equiv 2\pmod{18}$. Taking mod 7, $7\mid 1+2\times 3^{n-2a}\implies n-2a\equiv 1\pmod{6}$. Contradiction! Now if $a>n-a$, $3^a+2\times 3^{n-a}=3^{n-a}(3^{2a-n}+2)$. Similarly, taking mod 19, we get $19\mid 3^{2a-n}+2\implies 2a-n\equiv 16\pmod{18}$ and taking mod 7 gives $2a-n\equiv 5\pmod 6$. Contradiction! So $(2,3),(2,5)$ are the only solutions for $(a,n)$, so $(m,n)=(9,3),(6,3),(9,5),(54,5)$.

Can someone explain again the part with the moduls,please?

Similar to oneplusone's solution, WLOG assume $m=3^t$. Therefore we get $3^t+2\cdot 3^{n-t}=2^{n+1}-1$. Suppose $2<t$ and $2<n-t$, we get$27|3^t+2\cdot 3^{n-t}=2^{n+1}-1$.So $18|n+1$, and $73|2^18-1|2^{n+1}-1$. But $3^i\equiv \pm(1,3,9,27,8,24)\small(mod73)$, and it is easy to check that there is no $i,j\in \mathbb{Z}$ such that $3^i+2\cdot 3^j\equiv 0\small(mod73)$.That's contradiction and we get $t=0,1,2$ or $n-t=0,1,2$. After some research with inequalities, we get $(n,t)=(3,2),(5,2)$. Therefore$(m,n)=(6,3),(9,3),(9,5),(54,5)$.

We have $m|2\cdot 3^n$ so $m=3^k$ or $m=2\cdot 3^k$. First case $m=3^k$. $3^{2k}+2\cdot3^n=3^k(2^{n+1}-1)$ $3^k+2 \cdot 3^{n-k}=(2^{n+1}-1)$ Taking $mod 4$ we have that $k=2m$ is even and $n+1$ is even. We have $3^{min(k,n-k)}|2^n+1$ so $min(k,n-k)\le V_3(n+1)+1$ so we have that $2^{n+1}-1=3^k+23^{n-k}>3^{max(n,n-k))}=3^{n-V_3(n+1)-1)}>3^{n-log_3(n+1)}=3^{n}/(n+1)$ which implies // $2^{n+1}(n+1)>3^{n}$ so $n+1\le 10$. If $m=23^k$: $23^k+3^{n-k}=2^{n+1}-1$ so we have $min(k,n-k)\le V_3(n+1)+1<log_3(n+1)+1$ but $2^{n+1}>2^{n+1}-1=23^k+3^{n-k}>3^{max(k,n-k)}>3^{n-min(k,n-k)}>3^{n-log_3(n+1)}=3^{n}/(n+1)$ so $n+1\le 10$. Working out each $n$ we get $(n,m)=(3,6),(3,9),(5,9),(5,54)$.

The answer is $(m,n) \in \left\{ (9,3), (6,3), (9,5), (54,5) \right\}$, which work. As the equation is a quadratic in $m$ for any fixed $n$, we will show that $n \notin \{3,5\}$ do not work. Taking the discriminant, it's equivalent to solving \[ t^2 = \left( 2^{n+1}-1 \right)^2 - 4 \cdot 2 \cdot 3^n \iff 8 \cdot 3^n = \left( 2^{n+1}-1 \right)^2 - t^2. \]The right-hand side factors as a difference of squares, where both factors sum to $2(2^{n+1}-1)$. In particular, these factors have the same parity, and thus must be of the form $2 \cdot 3^a$ and $4 \cdot 3^b$, for nonnegative integers $a$, $b$ with $a+b=n$. Rearranging, we obtain \[ 2^{a+b+1} - 1 = 3^a + 2 \cdot 3^b. \]We have solutions $(a,b) = (2,1)$ and $(a,b) = (2,3)$, which corresponds to $n=3$ and $n=5$ above. Manual inspection reveals no other solutions with $\min(a,b) \le 2$, so we will suppose $\min(a,b) > 3$ and derive a contradiction. Taking modulo $8$, we find $a$ is even and $b$ is odd (hence unequal). The idea is that the exponent of $3$ is much too large in the above equation. Indeed, \begin{align*} \min(a,b) &\le \nu_3(3^a + 2 \cdot 3^b) = \nu_3\left( 2^{a+b+1}-1 \right) \\ &= \nu_3\left( 4^{\frac{a+b+1}{2}} - 1 \right) = 1 + \nu_3\left( \frac{a+b+1}{2} \right) \\ &\le 1 + \log_3\left( \frac{a+b+1}{2} \right) \le 1 + \log_3(\max\{a,b\}) \\ \implies \max(a,b) &\ge 3^{\min(a,b)-1}. \end{align*}Intuitively, this is way too big. We write \begin{align*} 2^{\max(a,b) + \min(a,b) + 1} &= 2^{a+b+1} = 1 + 3^a + 2 \cdot 3^b > 3^{\max(a,b)} \\ \implies 2^{\min(a,b)+1} &> \left( \frac 32 \right)^{\max(a,b)} \ge \left( \frac 32 \right)^{3^{\min(a,b)-1}} \end{align*}which is false for $\min(a,b) \ge 3$, as desired.

oneplusone wrote: I have another competely different solution. Taking mod $m$, we get $m\mid 2\times 3^n$. Since if $(m,n)$ is a solution, $\left(\frac{2\times 3^n}{m},n\right)$ is also a solution, so WLOG assume $m=3^a$. Then we have $3^a+2\times 3^{n-a}=2^{n+1}-1$. If either $a<3$ or $n-a<3$, it is easy to check that $(a,n)=(2,3),(2,5)$ are the only solutions by simple bounding. So now assume both $a$ and $n-a$ are at least 3. Taking mod 27, we get $18\mid n+1$. If $a\leq n-a$, then $3^a+2\times 3^{n-a}=3^a(1+2\times 3^{n-2a})$. Taking mod 19, we get $19\mid RHS$, so $19\mid 1+2\times 3^{n-2a}\implies n-2a\equiv 2\pmod{18}$. Taking mod 7, $7\mid 1+2\times 3^{n-2a}\implies n-2a\equiv 1\pmod{6}$. Contradiction! Now if $a>n-a$, $3^a+2\times 3^{n-a}=3^{n-a}(3^{2a-n}+2)$. Similarly, taking mod 19, we get $19\mid 3^{2a-n}+2\implies 2a-n\equiv 16\pmod{18}$ and taking mod 7 gives $2a-n\equiv 5\pmod 6$. Contradiction! So $(2,3),(2,5)$ are the only solutions for $(a,n)$, so $(m,n)=(9,3),(6,3),(9,5),(54,5)$. why did you just assume $m=3^a$ ??? How about $m=2.3^a$ ??? Because it said that $m\mid 2\times 3^n$ Please, I don't understand. Thanks

cjfev4 wrote: oneplusone wrote: I have another competely different solution. Taking mod $m$, we get $m\mid 2\times 3^n$. Since if $(m,n)$ is a solution, $\left(\frac{2\times 3^n}{m},n\right)$ is also a solution, so WLOG assume $m=3^a$. Then we have $3^a+2\times 3^{n-a}=2^{n+1}-1$. If either $a<3$ or $n-a<3$, it is easy to check that $(a,n)=(2,3),(2,5)$ are the only solutions by simple bounding. So now assume both $a$ and $n-a$ are at least 3. Taking mod 27, we get $18\mid n+1$. If $a\leq n-a$, then $3^a+2\times 3^{n-a}=3^a(1+2\times 3^{n-2a})$. Taking mod 19, we get $19\mid RHS$, so $19\mid 1+2\times 3^{n-2a}\implies n-2a\equiv 2\pmod{18}$. Taking mod 7, $7\mid 1+2\times 3^{n-2a}\implies n-2a\equiv 1\pmod{6}$. Contradiction! Now if $a>n-a$, $3^a+2\times 3^{n-a}=3^{n-a}(3^{2a-n}+2)$. Similarly, taking mod 19, we get $19\mid 3^{2a-n}+2\implies 2a-n\equiv 16\pmod{18}$ and taking mod 7 gives $2a-n\equiv 5\pmod 6$. Contradiction! So $(2,3),(2,5)$ are the only solutions for $(a,n)$, so $(m,n)=(9,3),(6,3),(9,5),(54,5)$. why did you just assume $m=3^a$ ??? How about $m=2.3^a$ ??? Because it said that $m\mid 2\times 3^n$ Please, I don't understand. Thanks Because if we get $m=3^a$ ———> $(m,n)$ is a solution,we will get $m’=\frac{2\times 3^n}{m}$ and $2|m’$ ,$(m’,n)$ is also a solution

We have \[ m^2-(2^{n+1}-1)m + 2\cdot 3^n = 0. \]So \[ (2^{n+1}-1)^2 - 8\cdot 3^n = k^2 \]for some $k$. Then \[ 8\cdot 3^n = (2^{n+1}-1-k)(2^{n+1}-1+k). \]So for some $a,b$ with $a+b=n$, we have \[ \{ 2^{n+1}-1-k, \ 2^{n+1}-1 + k \} = \{2\cdot 3^a,\ 4\cdot 3^b \}.\]We can't have $3^a, 8\cdot 3^b$ because then the sum of the two would be odd, but it is clearly even. Now, adding gives \[ 2^{n+2} - 2 = 2\cdot 3^a + 4\cdot 3^b \implies 2^{n+1} - 1 = 3^a + 2\cdot 3^b. \]mod 8 gives $a$ is even and $b$ is odd, so $n=a+b$ is odd too. Hence $n+1$ is even, so \[ \nu_3(2^{n+1}-1) = \nu_3(2^{n+1} - (-1)^{n+1}) = 1 + \nu_3(n+1) \ge \min(a,b). \]Then \[ 1+\nu_3(a+b+1) \ge \min(a,b) \implies a+b+1 \ge 3^{\min(a,b)-1}. \]This is a strong bound. We also know $2^{a+b+1} - 1 = 3^a + 2\cdot 3^b$. Intuitively, the RHS is way too big, using the above bound. Suppose $b>a$, in order to make the RHS bigger. We will still use size to narrow down the possibilities. So $\min(a,b)=a, \max(a,b)=b$. We know $a+b+1 \ge 3^{a-1}$, so $b\ge 3^{a-1}-a-1$. Then \[ 2^{a+b+1} = 1+3^a+2\cdot 3^b > 3^b \implies 2^{a+1} > 1.5^b \ge 1.5^{3^{a-1}-a-1}. \]If $a\ge 4$, this is not true. So $a\le 3$. Now manual case check. It turns out the answer is s $(m,n) =(9,3), (6,3), (9,5), (54,5)$.

View the equation as a quadratic in $m$; then by Vieta there must be integers $x,y$ with $xy=2 \cdot 3^n$ and $x+y=2^{n+1}-1$. WLOG let $x$ be even, and let $x=2 \cdot 3^a, y=3^b$; then we want to solve the equation $$2 \cdot 3^a+3^b = 2^{a+b+1}-1$$. Take $\nu_3$ of both sides to get by LTE that $$\text{min}(a,b) \leq \nu_3 \left( 4^{\frac{a+b+1}{2}}-1 \right) = 1+\nu_3(a+b+1)<1+\log_3(a+b+1)$$which implies that $$a+b+1 > 3^{\text{min}(a,b)-1}$$. For convenience, let $\text{min}(a,b)=t$ and $\text{max}(a,b)=T$. Now note that $2^{a+b+1}>3^{T}$ so from the previous centered equation we have $$2^{t+1}>1.5^T=1.5^{a+b-t}>1.5^{3^{t-1}-1-t}$$which is only true when $t \leq 3$. From here it is easy to manually check that the only solutions are $$(m,n)=\boxed{(6,3),(9,3),(9,5),(54,5)}$$.

Amir Hossein wrote: Find all pairs $(m,n)$ of nonnegative integers for which \[m^2 + 2 \cdot 3^n = m\left(2^{n+1} - 1\right).\] Proposed by Angelo Di Pasquale, Australia $$m|2\cdot 3^n \implies m=3^k ; m=2\cdot 3^k$$after several definitions, both of these cases are sufficient to solve the following equation. $$2\cdot 3^a + 3^b=2^{a+b+1}-1$$1)$a=b$ in which case it is not difficult to solve.(mod8). 2)$a>b \implies 3^b|2^{a+b+1}-1 \implies v_3(a+b+1) \ge b-1 \implies a+b+1 \ge 3^{ v_3(a+b+1)} \ge 3^{b-1}$ $$2^{a+b+1} > 2 \cdot 3^a$$$$(\frac{3}{2})^{2b} > 2^b > (\frac{3}{2})^a \implies 2b>a >3^{b-1}-b-1$$this leads to a boundary of $b$, and it is easy to see the rest, just as $b> a$.

Note that $m\mid 2\cdot 3^n$. Also, if $m$ is a solution for a fixed $n$ then $\frac{2\cdot 3^n}{m}$ is also a solution. Therefore we may WLOG $m=2\cdot 3^k$. Note that we then get $$2^{n+1}-1=3^{n-k}+2\cdot 3^k.$$Now if $k=0$ or $k=n$ we can check that no solutions exist which implies that $2^{n+1}\equiv 1\pmod 3$ or that $n$ is odd. Then we have \[\nu_3(2^{n+1}-(-1)^{n+1})=1+\nu_3(n+1)=\text{min}(k,n-k)\]so now we split into cases. If $k> \frac{n}{2}$ then we have $\nu_3(n+1)=n-k-1$ which implies that $3^{n-k-1}\le n+1$ or that $3^k\ge \frac{3^{n-1}}{n+1}$ which means that \[2^{n+1}-1=3^{n-k}+2\cdot 3^k>\frac{2\cdot 3^{n-1}}{n+1}\]so that $n\le 8$. Then $n=1,3,5,7$ which gives $k=0,2,3,6$. Only $(k,n)=(3,5)$ works and gives the solutions $(m,n)=(54,5),(9,5)$. Now suppose that $k<\frac{n}{2}$. Then we have $\nu_3(n+1)=k-1$ so we get $3^{k-1}\le n+1$ and $3^{n-k}\ge \frac{3^{n-1}}{n+1}$. Bounding, we get $n\le 10$, so that $n=1,3,5,7,9$ and $k=1,1,2,1,1$. Only $(k,n)=(1,3)$ works so we get the final two solutions $(m,n)=(6,3),(9,3)$. We are done. $\blacksquare$

We have $m(2^{n+1}-1-m)=2\cdot 3^n$, so $\{m,2^{n+1}-1-m\}=\{x,y\}$ where $x=2\cdot 3^i$ and $y=3^j$ and $x+y=2^{i+j+1}-1.$ Consider $\nu_3(x+y)=\min(i,j).$ If $2\nmid n+1=i+j+1$ the RHS isn't divisible by $3$, so $i=0$ or $j=0.$ Thus, $(x,y)=(2,3^n)$ or $(1,2\cdot 3^n).$ One of these equals $m$ so pluggin $m=1,2,3^n,2\cdot 3^{n}.$ $m=1$ gives no solutions by bounding, For $m=2$, take mod $3$ to get $0\equiv 2^{n+2}\pmod 3$, absurd. For $m=3^n$ does not work again because we get $3^n+3=2^{n+1}$ so $3\mid 2^{n+1}.$ For $m=2\cdot 3^{n}$ does not work by bounding. In summary, no solutions. Now, we have $2\mid n+1$ so \[\min(i,j)=\nu_3(4^{\frac{n+1}{2}}-1)=1+\nu_3(\frac{n+1}{2})=1+\nu_3(i+j+1)\le 1+\log_3(i+j+1)\]Thus, $i+j+1\ge 3^{\min(i,j)-1}$ Now, $2^{n+1}\ge 3^{\max(i,j)}$ so $2^{\min(i,j)+1}\ge 1.5^{\max(i,j)}=1.5^{i+j-\min(i,j)}>1.5^{3^{\min(i,j)-1}-1-\min(i,j)}.$ Suppose $\min(i,j)>3$ then this is a contradiction. Now, check $\min(i,j)=1,2,3$ to get our solutions.

Nice problem ... My solution is quite long so I will share in short way with some parts to solve it. Part 1 Taking mod $m\implies$ $m\mid 2\times 3^n$ Part 2 Let $m=3^a$ Part 3 We show that if $a>2$ and $n-a>2\implies$ $n\equiv -1\pmod{18}$ (by mod$27$ ) Part 4 Then $19\mid 2\times 3^a + 3^{n-a}$ Part 5 Divide into two cases $n>2a$ or $n<2a$ Part 6 Prove that there is not any solution in both cases(from mod$19$ and mod$2$) Part 7 Thus $a>2$ and $n-a>2$ can not be true at same time , Work on $1) a=1 , 2) a=2 , 3) a>2,n-a<2$ Part 8 Find $m=9 , n=3$ and $m=9 , n=5$ Part 9 Now let $m= 2\times 3^a$ and do same things above Part 10 Find $m=6 , n=3$ and $m=54 , n=5$ So we are done.

Take the equation as a quadratic in $m$. Since the discriminant is a square, we have \[ (2^{n+1} - 1)^2 - 8\cdot 3^n \implies 8\cdot 3^n = (2^{n+1} - 1 + k)(2^{n+1} - 1 - k)\]Note that the factors must be of the form $2\cdot 3^x$ and $4\cdot 3^y$, with $x + y = n$. Allowing $k$ to be negative, we may assume WLOG that the first factor is $2\cdot 3^x$ and the second one is $4\cdot 3^y$. Hence, \[k = 3^x - 2\cdot 3^y \text{ and } 2^{n+1} - 1 = 3^x + 2\cdot 3^y\]We now bound $x$ and $y$. We have \[3^x < 2^{n+1} \implies 3^y > \frac{3^n}{2^{n+1}} \text{ and } 3^y < 2^n \implies 3^x> \frac{3^n}{2^n}\]Hence, \[\frac{3^n}{2^{n+1}} < 3^{\min (x, y)} < 3^{\nu_3 (2^{n+1} - 1)} = 3^{1 + \nu_3 ((n+1)/2)} \le 3\cdot \frac{n+1}{2}\]This gives $n\in \{1, 3, 5, 7\}$. We get the solutions $\boxed{(m, n) = (6, 3), (9, 3), (9, 5), (54, 5)}$.

shows that $m=1$ and $m=2$ lead to no solutions, which implies that $3\mid m_{1}$ and $3\mid m_{2}$, so $3\mid m_{1}+m_{2} = 2^{n+1}-1$, so $n+1$ is even. WLOG let $m_{1}\leq m_{2}$. Now $m_{1}m_{2}=2\cdot 3^{n}\Longrightarrow \nu_{3}(m_{1})\leq \nu_{3}(m_{2})$ whence

we have that $\sqrt{2\cdot 3^{n}}>\frac{3(n+1)}{2}$, so we have that:

This implies that $n\leq 5$ and we're left to check the remaining cases. In the beginning, we mentioned that $n$ must be odd, so $n\in\{1,3,5\}$. A direct check shows that $n=1$ doesn't lead to a solution whereas $n=3$ and $n=5$ lead to the solutions $(m,n)=(6,3), (9,3), (9,5), (54,5)$.

Note that $m \mid 2 \cdot 3^n$. Assume first that $m = 2 \cdot 3^k$ for some $k \leq n$. Then, the equation simplifies to $$2 \cdot 3^{2k} + 3^n = 3^k(2^{n+1} - 1).$$We take $\nu_3$ of both sides. If $\nu_3(2^{n+1} - 1) = 0$, then we must have $n = k$. This reduces to $2 \cdot 3^n + 1 = 2^{n+1} - 1$, which is obviously impossible for size issues. Otherwise, $n$ must be odd, and set $r = \nu_3(2^{n+1} - 1) = 1 + \nu_3\left(\frac{n+1}2\right) \leq 1+\log_3(n+1) - \log_3(2)$. Then if $r = n-k$, then $$2 \cdot 3^k + 3^{n-k} = 2^{n+1} - 1.$$But on the other hand $$n-k \leq \log_3(n+1) + 1 \iff n+1 \geq 3^{n-k-1}.$$But this implies $3^k \geq \frac{3^{n-1}}{n+1}$, or $$\frac{3^{n-1}}{n+1} \leq 2^{n+1} - 1 \iff n \leq 10.$$Finite case check yields $(6, 3)$ and $(54, 5)$. The other case works out in the exact same way by symmetry. Now, if $m =3^k$, notice that the transformation $m \to \frac{2 \cdot 3^n}m$ bijects between solutions with $m = 2 \cdot 3^k$ and $m = 3^k$. Thus, the other two solutions are $(9, 5)$ and $(9, 3)$.

It is easy to see $m\mid 2\cdot 3^n$, and if $\left(n, m\right)$ is a solution, then so is $\left( n, \frac{2\cdot 3^n}{m}\right)$, WLOG, $m=2\cdot 3^k$. We can rewrite the equation is $2\cdot 3^k + 3^{n-k} = 2^{n+1}-1$. If $k<3$ or $n-k<3$, we can easily check that $(n,m) = (5,9), (5,54), (3,9), (3,6)$ are the only solutions. Taking mod 27, we have $n\equiv 17 \pmod{18}$, but taking mod 73, we get $2\cdot 3^k + 3^{n-k} \equiv 0 \pmod{73},$ but $3^n \bmod{73}$ is either $3^5$ or $3^{11}$ (since the order of $3$ modulo $73$ is 12), and it is easy to check none of these yield solutions. Motivation: I found some primes that divided $2^{18}-1$ and $73$ was interesting, since the order of 3 was also nice. From that it was fairly obvious.

:skull: i only found the pairs (6,3),(9,3),(9,5) after a hour of dumb bashing

Same as most of the other ones, but this is a pretty nice problem. First, mod m, we get $m\mid 3^n2$; in particular, we claim n is odd: If n were even note that 2^{n+1}-1 is not 0 mod 3. mod 3 we get $m^2\equiv m\implies m\equiv \{0,1\}\pmod 3$, so there are two cases. 1. m is 0 mod 3. Looking at $v_3, m=\{3^n,3^n2\}$. If $m=3^n,3^n(3^n+2)=3^n(2^{n+1}-1)$ has no solutions. If $m=3^n2, 3^n2(3^n2+1)=3^n2(2^{n+1}-1)$ also has no solutions. 2. m is 1 mod 3 from here we get a contradiction because we derive m=1, and checking, there are no sols. We conclude that n is odd, and 2^n-1 is not divisible by 3. Now, check manually the cases for n<10. Looking at the discriminant in the quadratic of m, we have $$(2^{n+1} - 1)^2 - 3^n8=k^2 \implies3^n8 = (2^{n+1} - 1 + k)(2^{n+1} - 1 - k)=ab\stackrel{\text{same mod 2,WLOG}}{\implies}(a,b)=(3^x2,3^y4),x + y = n\implies(k,2^{n+1}-1) = (3^x - 3^y2,3^x + 3^y2)\quad(1); k<2^{n+1}-1\implies 3^x2<2^{n+1}2-1\implies 3^x<2^{n+1}\implies 3^y > \frac{3^n}{2^{n+1}}\quad(2),3^y4<2^{n+1}\implies3^y < 2^{n-1} \implies 3^x> \frac{3^n}{2^{n+1}}\quad(3)$$$$\stackrel{(2),(3)}{\implies} \frac{3^n}{2^{n+1}} < 3^{\min (x, y)} \stackrel{(1)}{=} 3^{\nu_3(2^{n+1}-1)}=3^{\nu_3 ((2^n - 1)(2^n+1))} =3^{\nu_3(2^n+1)}= 3^{1 + \nu_3n} \le 3n\implies n\le10\implies(m, n) = \{(6, 3), (9, 3), (9, 5), (54, 5)\},$$as desired. $\blacksquare$ Remark. The part needing n is odd might be thought of as an unnecessary procedure, but it is well motivated when we get to the last few steps to need to derive it.

We can take the equation $\pmod{m}$ to see that $2 \cdot 3^n \equiv 0 \pmod{m}$. We also know that $m$ must be odd because taking $v_2$, thus $m = 3^a$ or $m = 2\cdot 3^a$ for some positive integer $a \leq n$. We first take care of $m = 3^a$ We can divide the equation by $3^a$ to get \[3^a + 2 \cdot 3^{n-a} = 2^{n+1} -1\]We can see that $v_3(3^a + 2 \cdot 3^n-a) \geq \min(a, n-a)$ (note the expression is not equal only when $2a =n$). We can see that the LHS is always divisible by $3$ (if $a = 0$ then the LHS is even which is not possible). This means that we can assume that $n+1$ is even. Thus we can see taking $v_3$ that we can see that we need \[\min(a, n-a) \leq 1+v_3\left(\frac{n+1}{2}\right)\]Notice that the first value of $n$ to attain $x$ on the LHS is $2 \cdot 3^{x-1} - 1$. This means that we need \[3^x + 2 \cdot 3^{2 \cdot 3^{x-1}-x - 1} \leq 2^{2 \cdot 3^{x-1}} - 1\]or \[3^{2 \cdot 3^{x-1}-x - 1} + 2 \cdot 3^x \leq 2^{2 \cdot 3^{x-1}} - 1\]The first equation obviously has size issues when $x \geq 3$. Checking those finite cases we get $(m, n) = (9, 5)$ Similarly we can see that the second equation has size issues when $x \geq 3$. Checking finite cases we get $(m, n) = (9, 3)$. We can see that if $m = 2 \cdot 3^a$ then we can divide by $m$ to get $2 \cdot 3^a + 2 \cdot 3^{n-a} = 2^{n+1} - 1$ This solution set is actually isomorphic to the $m = 3^a$ solutions. Which give $\boxed{(m, n) = (9, 5), (9, 3), (54, 5), (6, 3)}$. $\blacksquare$

What in the world is my solution. We claim the only solutions are $(m,n) \in \left\{ (9,3), (6,3), (9,5), (54,5) \right\}$. Taking modulo $m$ we have $m \mid 2 \cdot 3^n$. Thus let $m = 2 \cdot 3^b$ with $1 \leq b \leq n$. Then we have, \begin{align*} 4 \cdot 3^{2b} + 2 \cdot 3^n &= 2^{n + 2}3^b - 2\cdot 3^b\\ 2 \cdot 3^{2b} + 3^n &= 2^{n+1}3^b - 3^b\\ 2 \cdot 3^b + 3^{n-b} &= 2^{n+1} - 1\\ 3^{n-b} + 1 &= 2^{n+1} - 2 \cdot 3^b \end{align*}Clearly $\nu_2(\text{RHS}) = 1$ and hence $n - b $ must be odd. Specifically $n$ and $b$ are of opposite parity. Reindex such that $n = a + b$ whence $a$ and $b$ must be of opposite parity. Then we have, \begin{align*} 3^a + 1 &= 2^{a + b + 1} - 2 \cdot 3^b\\ 2^{a+b+1} - 1 &= 3^a + 2 \cdot 3^ b \end{align*}For $a, b \leq 3$ we can find solutions $n = 3$, $n = 5$. Assume that $\min(a, b) > 3$. Taking modulo $2 \cdot 3^3$ and modulo $3^3$ we find that $a \equiv 0 \pmod{18}$ and $b \equiv -1 \pmod{18}$. Finally taking modulo $127$ we find that, \begin{align*} 0 &\equiv 3^a + 2 \cdot 3^b \pmod{127}\\ -3^a &\equiv 2 \cdot 3^b \pmod{127}\\ 3^{a - b} &\equiv 125 \pmod{127} \end{align*}Now we claim that there are no solutions. Indeed looking at the residues $3$ leaves for powers congruent to $1$ modulo $18$ we see, $3^{18k + 1} \in \{3, 12, 48, 65, 6, 24, 96\}$ hence we are done.

Clearly $m\mid 2\cdot 3^n$ We are going to check the case when $m=2\cdot 3^\theta$ then $$4\cdot 3^\theta+2\cdot 3^{n-\theta}=2^{n+1}-1$$But this fails by $\pmod 2$ So $m=3^k$ and the equation becomes $$3^k+2\cdot 3^{n-k}=2^{n+1}-1$$We can easily check that nor $k=0, n=k$, then $n$ must be odd Now check $\nu_3$ of the equation and note that $$\min\{k,n-k\}=\nu_3(3^k+2\cdot 3^{n-k})=\nu_3\left(4^{\frac{n+1}{2}}-1\right)=1+\nu_3\left(\frac{n+1}{2}\right)\leq 1+\log_3\left(\frac{n+1}{2}\right)\leq 1+\log_3\left(\max\{k,n-k\} \right)$$so $\max\{n,n-k\} \geq 3^{\min\{k,n-k\}}$ But in the equation $$2^{\max\{n,n-k\}+{\min\{k,n-k\}}}=3^k+2\cdot 3^{n-k}+1>3^{\max\{n,n-k\}}$$so $$2^{\min\{k,n-k\}+1}>\left(\frac{3}{2} \right)^{\max\{n,n-k\}}\geq \left(\frac{3}{2}\right)^{3^{\min\{k,n-k\}}} $$Which fails for ${\min\{k,n-k\}}\geq 3$ so we can manually check for ${\min\{k,n-k\}}=1,2,3$

We claim our only solutions are $\boxed{(9,3), (6,3), (9,5), (54,5)}$. Expressing the equation as a quadratic in $m$, we require \[8 \cdot 3^n = (2^{n+1}-1)^2-x^2 = (2^{n+1}-1-x)(2^{n+1}-1+x)\] for an integer $x$. Parity tells us these factors correspond to $(2 \cdot 3^y) \cdot (4 \cdot 3^{n-y}$. Modulo 8 further tells us $y \neq n-y$, so assume $y<n-y$ (where the other case is analogous) to give \[2^{n+1}-1 = 2 \cdot 3^y + 3^{n-y} = 3^y \left(3^{n-2y}+2\right).\] If $y$ or $n-y$ are at most 2, casework reveals the solutions above. Otherwise, $n+1$ must be even, so we use LTE to find \[v_3(2^{n+1}-1) = v_3\left(\frac{n+1}{2}\right)+1 \ge y \implies n+1 \ge 2 \cdot 3^{y-1},\] which is evidently too large if $y \ge 3$. (One way to see this is by noting the RHS of our equation grows faster than the left, so we can isolate $n$ and find an upper bound in terms of $y$ which excludes the interval we just proved.) $\blacksquare$

Quadratic formula forces that $(2^{n +1} - 1)^2 - 8 \cdot 3^n$ is a perfect square, using difference of squares we get $(2^{n + 1} - 1 - k )\cdot (2^{n + 1} - 1 + k) = 8 \cdot 3^n$, so one of the two terms on the left side is $2 \cdot 3^a$, the other is $4 \cdot 3^b$, so we get $2 \cdot 3^a + 4 \cdot 3^b = 2(2^{n+ 1} - 1)$, so we get $3^a + 2 \cdot 3^b = 2^{n + 1} - 1$, with $a + b = n$. If $n$ is even, the right side is not divisible by $3$, so one of $a,b$ is $0$. If $a = 0$, we get $3^n = 2^n - 1$, which obviously has no solutions. If $b = 0$, we get $3^n = 2^{n + 1} - 3$., which also has no solutions by size. Thus $n$ is odd, so $\nu_3(2^{n + 1} - 1) = 1 + \nu_3(\frac{n +1}{2})$, and $\nu_3(3^a + 2 \cdot 3^b) \ge \min(a,b)$, so $1 + \log_3 n \ge 1 + \log_3 \frac{n + 1}{2} \ge 1 + \nu_3(\frac{n +1}{2}) = \nu_3(3^a + 2 \cdot 3^b) \ge min(a, n - a)$. Thus $\max(a,b) \ge n - 1- \log_3 n$. Thus we can bound $2^{n + 1 } - 1 =3^a + 2 \cdot 3^b \le 3 \cdot 3^{\max(a,b)} \le 3^{n - \log_3 n} = \frac{3^n}{n}$. Thus, we can start by eliminating all $n$ for which $3^{n} > (2n)2^n$, or equivalently $(\frac{3}{2})^n \ge 2n$. We can observe $\frac{3^7}{2^7} = \frac{2187}{128} > 14$, then taking derivatives of both sides, we see $(\frac 32)^n \ln \frac 32 > 2$ for $n \ge 7$ as $\ln 32 < 3$ as $(1.5)^3 > 3 > e$. We then manually solve the cases for $n$ between $0$ and $6$. We already forced $n$ odd, so it remains to check $n = 1,3,5$. Checking $n = 1$ gives $m^2 -3n + 6$, which has no solutions, checking $n = 3$ gives $m^2 + 54 = 15m$, which resolves to $(m - 6)(m- 9) =0$, so we have solutions $(6,3), (9,3)$, both of which clearly work. Checking $n = 5$ gives $m^2 + 486 = 63$, which resolves to $(m - 54)(m - 9) = 0$, so we have solutions $(54, 5), (9,5)$, both of which clearly work. We have exhausted all cases, so we are done.

Solved a while ago but forgot to post The only solutions are \[(m,n) = (6,3), (9,3), (9,5), (54,5), \]which clearly work. Now we show they are the only ones If $n=3$, then $m^2 - 15m + 54 = 0\implies m\in \{6,9\}$. If $n=5$, then $m^2 - 63m + 486 = 0\implies m\in \{9,54\}$. Thus it suffices to show that $n\in \{3,5\}$. Taking the equation as a quadratic in $m$, by taking the discriminant, we find \[k^2 = (2^{n+1} - 1)^2 - 8\cdot 3^n \]for some nonnegative integer $k$, so \[8\cdot 3^n = (2^{n+1} - k - 1)(2^{n+1} + k - 1)\] Since the two terms in the product are of the same parity, they are equal to $2\cdot 3^a$ and $4\cdot 3^b$ in some order. So $n=a+b$. The equation is now equivalent to \[2\cdot 3^a + 4\cdot 3^b = 2^{a+b+2}- 1,\]so \[3^a + 2\cdot 3^b = 2^{a+b+1} - 1\] If $a=0$, then $2\cdot 3^b + 1 = 2^{b+1} - 1$, absurd. If $b=0$, then $3^a + 2 = 2^{a+1} - 1$, also absurd. Now taking $\pmod 8$ gives $a$ even and $b$ odd (so $a\ne b$). Since $a+b$ is odd, we have $\nu_3(2^{a+b+1} - 1) = \nu_3(a+b+1) + 1$ (this can be found by factoring and LTE). Let $\min(a,b)= c$ and $\max(a,b) = M$. Taking $\nu_3$ of the equation, we get \[c-1 = \nu_3(a+b+1) .\]Since $a+b+1$ is even, we have \[a+b+1 \ge 2\cdot 3^{c - 1}\] On the other hand, we get \[2^{a+b+1} > 3^{M}\]so \begin{align*} 2^{c + 1} \\ > (1.5)^{M}\\ >M \\ \end{align*} We have \[2M\ge a+b+1,\]so \[2^{c+1} > 3^{c- 1} \implies c\le 4\] Also \[2^{M + 5} \le 2^{c + M + 1} >3^M, \]so $M\le 8$. If $c=4$, then since $b$ is odd, we have $a=4$. So \[81 + 2\cdot 3^b = 2^{b+5} - 1\]Its easy to check that no odd $b\le 8$ satisfies this. If $c=3$, then since $a$ is even, we have $b=3$. So \[3^a + 54 = 2^{a+4} - 1\]It's easy to check that all even $a\le 8$ do not satisfy this. If $c=2$, then since $b$ is odd, we have $a=2$. So \[9 + 2\cdot 3^b = 2^{b+3} - 1\]Here we can check that the only possibility is $b=3$, since $b>c = 2$. This gives $n=5$. If $c=1$, then since $a$ is even, we have $b=1$. So \[3^a + 6 = 2^{a+2} - 1\]Here we must have $a=2$, which gives $n=3$. Thus $n\in \{3,5\}$.