Perfect

If $n^2+12n+65=\left(n+6\right)^2+29=k^2$ for some integer $k$ , then $(k+n+6)(k-n-6)=29$ . Since 29 is a prime number, hence $\begin{cases} k+n+6=a \\ k-n-6=b \end{cases}$ where $(a,b)=(29,1), (1,29), (-29,-1), (-1,-29)$ . The difference of the two equations gives $n=8, -20, -20$ and $8$ respectively. As the question requires for positive integer $n$ , we conclude that $n=8$ is the only answer.

By completing the square, we get

${ n }^{ 2 }+12n+65={ \left( n+6 \right) }^{ 2 }+29$

Through analysis, it can be found that difference of square number is always a unique odd number (in a particular order) -

$={ n }^{ 2 }-{ \left( n-1 \right) }^{ 2 }\\ =\left( n+n-1 \right) \left( n-n+1 \right) \\ =2n-1$

Now we want to find two squares numbers whose difference is 29. In other words, we need to find a positive integer $m$ such that ${ n }^{ 2 }+29={ m }^{ 2 }$ .

We can find this number using the previously discovered formula.

$2n-1=29\\ n=15$

Hence our numbers are 14 and 15. So $15^2 - 14^2 = 29$ . In our previous formula ${ \left( n+6 \right) }^{ 2 }+29$ , we find out that $n+6=14\Longrightarrow n=8$ .

---

Also a negative solution, $n = -20$ exists but we can ignore it since the questions only asks for positive numbers.

Uniqueness of the solution can be proved from the fact that ${ n }^{ 2 }-{ \left( n-1 \right) }^{ 2 }$ is unique for every distinct $n$ . Additionally, ${ \left( n+6 \right) }^{ 2 }+29$ cannot be factored further over reals.