If $a_{n+1}=\frac {a_n^2+5} {a_{n-1}}$ then $a_{n+1}=Sa_n+Ta_{n-1}$ for some $S,T\in \Bbb Z$. Question

Let $$a_{n+1}:=\frac {a_n^2+5} {a_{n-1}},\, a_0=2,a_1=3$$
Prove that there exists integers $S,T$ such that  $a_{n+1}=Sa_n+Ta_{n-1}$.

Attempt
I calculated the first few values of $a_n$: $a_2=7,a_3=18, a_4=47$ so I'd have the system of diophantine equations:
$$
7=3S+2T\\
18=7S+3T\\
47=18S+7T
$$
Now: it seems that all of the $a_i$ are pairwise coprime, so these equations should always have solutions, but how could I check that the intersection of all the solutions is not $\emptyset$?

From the initial data and recursion, compute $a_0=2,a_1=3,a_2=7,a_3=18$.
If we have an $S$ and $T$ so that $a_{n+1}=Sa_n+Ta_{n-1}$, then
$$
\begin{bmatrix}3&2\\7&3\end{bmatrix}\begin{bmatrix}S\\T\end{bmatrix}=\begin{bmatrix}7\\18\end{bmatrix}\implies\begin{bmatrix}S\\T\end{bmatrix}=\begin{bmatrix}3&2\\7&3\end{bmatrix}^{-1}\begin{bmatrix}7\\18\end{bmatrix}=\begin{bmatrix}3\\-1\end{bmatrix}
$$
Now, to verify that the linear recursion $a_{n+1}=3a_n-a_{n-1}$ implies
$$
a_{n+1}a_{n-1}-a_n^2=5\tag{1}
$$
Assume the linear recursion, then
$$
\begin{align}
&\left(a_{n+2}a_n-a_{n+1}^2\right)-\left(a_{n+1}a_{n-1}-a_n^2\right)\\
&=(3a_{n+1}-a_n)a_n-a_{n+1}^2-a_{n+1}a_{n-1}+a_n^2\\
&=3a_{n+1}a_n-a_{n+1}^2-a_{n+1}a_{n-1}\\
&=a_{n+1}(3a_n-a_{n-1})-a_{n+1}^2\\
&=a_{n+1}^2-a_{n+1}^2\\
&=0\tag{2}
\end{align}
$$
Thus, the linear recursion and $(1)$ imply that
$$
a_{n+2}a_n-a_{n+1}^2=5\tag{3}
$$
