This problem was proposed by Cyprus

The seqence is -5,1,9,19,31,45 From here we see the pattern of adding 6 then 8 then 10 and so on and to prove just write the formula for such series and make sure it is indeed the given one.

As @above said, it is easy to show by induction that $a_n = n^2 + 7n + 1$. Then we want $n^2 + 7n + 1 = m^2 \iff 4n^2 + 28n + 4 = (2m)^2$. For $n > 8$, we have $(2n+6)^2 < (2m)^2 = 4n^2 + 28n + 4 < (2n+7)^2$ so we only need to check for $1\leq n \leq 8$. After checking those, we get that $n=1,8$ are the only solutions.

hakN wrote: Then we want $n^2 + 7n + 1 = m^2 \iff 4n^2 + 28n + 4 = (2m)^2$. The motivation for multiplying by $4$ lies in the fact that we want to obtain the square of a binomial where all the terms are integers. Further, it could be completed by applying the difference of squares: $$(2n+7)^2-45=(2m)^2\text{,}$$$$(2n+7-2m)(2n+7+2m)=45\text{.}$$Since $2n+7-2m<2n+7+2m$, we have 3 cases: 1) $2n+7-2m=1$, $2n+7+2m=45\Rightarrow \boxed{n=8}$, $m=11$; 2) $2n+7-2m=3$, $2n+7+2m=15\Rightarrow \boxed{n=1}$, $m=3$; 3) $2n+7-2m=5$, $2n+7+2m=9\Rightarrow n=0$, $m=1$.

Note that the recursion equation rearranges to $(n+3)(a_{n+1}-a_n)=(n+4)(a_n-a_{n-1})$ this means that the first difference $\Delta a_n=a_{n+1}-a_n=4K\prod_{k=1}^n\frac{k+4}{k+3}=K(n+4)$ is an arithmetic sequence and so $a_n$ is quadratic. $a_n=An^2+Bn+C\implies A(n+1)^2+B(n+1)+C=\frac{(n+5)(An^2+Bn+C)+22}{n+3}\implies(7A-B)n+(3A+3B-2C-22)=0$ This means $\{B=7A,C=12A-11\}\implies a_n=A(n^2+7n+12)-11$. When $a_1=9$ we have $A=1$ so that $a_n=n^2+7n+1$ When are they square? $(n+4)^2-a_n=n+15>0$ $a_n-(n+3)^2>0\implies n>8$ So we need to check $1\leq n\leq 8$ and $\{a_1,\cdots,a_8\}=\{9,19,31,45,61,79,99,121\}\implies n\in\{1,8\}$.