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.

$(x-1)$ and $(x+1)$ must be power of 3. Then $x-1=3^a$ and $x+1=3^b$
$\Rightarrow 3^b-3^a = 2 \Rightarrow 3^a(3^{b-a}-1)=2$.
So $a=0$ and $b-a=1 \Rightarrow b=1$. Thus $x-1 = 3^a = 1$ and $x+1 = 3^b = 3$.
$\Rightarrow 3^k=1\cdot 3=3 \Rightarrow k = 1 \Rightarrow n=k+1=5$.
Therefore, the unique solution is $n=5$ with $3^5+81=324 = 18^2$.
