Sequence in which adding 2 produces a square Consider the sequence defined by $x_1=2$, $x_2=x_3=7$, $x_{n+1}=x_{n}x_{n-1}-x_{n-2}$.
Then $x_4=7 \cdot 7 - 2 = 47, x_5=47 \cdot 7 - 7=322, x_6=322 \cdot 47 - 7 =15127, x_7=15127 \cdot 322 -47=4870847,x_8=4870847 \cdot 15127-322=73681302247.$
My spreadsheet cannot calculate any more, but notice that $x_n+2$ is always square:
$\sqrt{x_4+2}=7$, $\sqrt{x_5+2}=18$, $\sqrt{x_6+2}=123$, $\sqrt{x_7+2}=2207$,$\sqrt{x_8+2}=271443.$ Is this a coincidence?

We have
\begin{align*} x_{n+1}+x_{n-2}&=x_{n}x_{n-1}\\
\end{align*}
then
\begin{align*} x_{n+1}x_{n-2}&=x_{n}x_{n-1}x_{n-2}-x_{n-2}^2\\
&=x_n(x_{n}+x_{n-3})-x_{n-2}^2\\
&=x_nx_{n-3}+x_{n}^2-x_{n-2}^2\\
\end{align*}
then by telescoping 
\begin{align*} x_{n+1}x_{n-2}&=x_4x_1+x_n^2+x_{n-1}^2-x_3^2-x_2^2\\
x_{n+1}x_{n-2}&=x_n^2+x_{n-1}^2-4\\
\end{align*}
Now we let $x_n=z_n-2$. We then have
\begin{align*}
z_{n+1}z_{n-2}-2z_{n+1}-2z_{n-2}+4&=z_n^2-4z_n+4+z_{n-1}^2-4z_{n-1}+4-4\\
z_{n+1}z_{n-2}&= z_n^2-4z_n+z_{n-1}^2-4z_{n-1}+2z_{n+1}+2z_{n-2}\\
z_{n+1}z_{n-2}&= z_n^2-4z_n+z_{n-1}^2-4z_{n-1}+2x_{n+1}+4+2z_{n-2}\\
\end{align*}
But \begin{align*}
x_{n+1}&=x_nx_{n-1}-x_{n-2}\\
&=(z_n-2)(z_{n-1}-2)-z_{n-2}+2
\end{align*}
then by substitution and reduction, we find
\begin{align*}
z_{n+1}z_{n-2}&=(z_n+z_{n-1}-4)^2
\end{align*}
from which a proof by induction easily follows.
Addition
We show that $v_n=\sqrt{\frac{x_{n}-2}{5}}$ is A101361 (shifted).
We have $z_n = 5v_n^2+4$ and  $x_n= 5v_n^2+2$. Then 
\begin{align*}
z_{n+1}z_{n-2}&=(z_n+z_{n-1}+4)^2\\
(5v_{n+1}^2+4)(5v_{n-2}^2+4)&=(5v_{n}^2+5v_{n-1}^2+4)^2
\end{align*}
which, after algebric manipulations, can be rearranged as 
$$5v_{n+1}^2v_{n-2}^2-5(v_{n}^2-v_{n-1}^2)=4\left(5v_{n}^2v_{n-1}^2 +2v_n^2 +2v_{n-1}^2 -(v_{n+1}^2+v_{n-2}^2)\right) \tag1 $$
On the other hand, we have
\begin{align*}
x_{n+1}+x_{n-2}&=x_nx_{n-1}\\
5v_{n+1}^2+2+5v_{n-2}^2+2&=(5v_{n}^2+2)(5v_{n-1}^2+2)
\end{align*}
which can be simplified so that 
$$v_{n+1}^2+v_{n-2}^2 = 5v_{n}^2v_{n-1}^2 +2v_n^2 +2v_{n-1}^2 \tag2$$
Comparing (1) and (2), we see that it must be that $v_{n+1}v_{n-2}=v_n^2 - v_{n-1}^2$ which is the third-order nonlinear recursion that is satisfied by A101361. Moreover, it is easy to verify that the initial terms are equal (shifted). Also it is said in that OEIS page that $v_{n+1}= F_{2F_{n}}$, where $F_{n}$ is the Fibonacci number. Then we finally obtain a nice closed form for the OP sequence, in terms of Fibonacci and Lucas numbers: 

$$ x_n = L_{2F_{n-1}}^2 -2$$

