<s>[INST] If $3^n+81$ is a perfect square, then positive integer value $n$ is If $3^n+81$ is a perfect square, Then calculation of  a positive integer value of $n$.
$\bf{My\; Try}::$ When $n≤4,$ then easy to know that $3^n+81$ is not a perfect square. 
Now let $n=k+4(k∈Z^{+}),$ then $3^{n}+81=81(3^{k}+1).$ 
So $3^{n}+81$ is a perfect square, and $81$ is square, 
there must be a positive integer $x$, such that $3^{k}+1=x^2⇒3^k=(x−1)⋅(x+1)$
Means $(x+1)$ and $(x-1)$ must be a power of $3$ form
Now I did not understand how can i solve after that
Help Required
Thanks.
 [/INST] Note that if $3^n+81 = x^2$, we have
$$(x+9)(x-9) = 3^n$$
Hence, we have
$x+9 = 3^m$ and $x-9 = 3^{n-m}$. Hence, we need two powers of $3$ that differ by $18$, i.e., we need $3^m - 3^{n-m} = 18$.
Now observe that $3^m-3^{m-1} \geq 18$, if $m\geq 3$. Hence, we have limited options for $m$ and $n-m$.
Hence, $m=3$ and $n-m=2$ and we are done.
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