Find all positive integer $m$ such that $3^m+81$ is a perfect square.
Attempt:
Let $3^m+81=s^2$ for a positive integer $s$. Then \begin{align*} 3^m+81=s^2 \iff 3^m = s^2-81 \iff 3^m = (s-9)(s+9). \end{align*} Hence, $s-9$ and $s+9$ are two positive integer which are powers of $3$ and differs by $18$. Let $s-9=3^a$ and $s+9=3^b$ for some positive integers $a$ and $b$. Then, $3^b-3^a=18$. Clearly, $a<b$ yielding $3^a \mid 3^b$, which means that $(s-9) \mid (s+9)$. This gives us $(s-9) \mid ((s+9)-(s-9)) = 18$. Hence, $(s-9) \mid 18$. The only positive factor of $18$ which is greater than one is $9 (=3^2)$. Thus, we obtained $3^a=9$, i.e., $a=2$. Since $3^b-3^a=18$, we have $3^b=18+9=27=3^3$ yields $b=3$. Therefore, the only positive integer solution $(a,b)$ satisfying $3^b-3^a=18$ is $(2,3)$, which gives us $s=18$. Now, $3^m+81=s^2$. Since $s=18$, then $3^m+81=18^2 \iff 3^m = 324-81=243=3^5 \implies m=5$.
Does the above approact correct? Many thanks in advanced.
WLOG assume $a \leq b$. Then $a + n = b$ for some $n$. $3^{b} - 3^{a} = 3^{a}(3^{n}-1) = 18$. $3^{n} - 1$ would need to divide $18$ for some $n$. You can easily check this $n$ can only be $1$. So $3^{a+1} - 3^{a} = 18$ gives $3^{a} = 9$ and so $a = 2$. This would lead to your solution and would be the only one.