Because $\big(a_{n-1}^2 +2\big) > 0$, therefore $a_n$ cannot equal $0$. Meanwhile, the given relation is equivalent to $$a_na_{n-2}-a_{n-1}^2 =2\implies a_{n+1}a_{n-1}-a_n^2 = 2$$ Therefore, $$a_na_{n-2}-a_{n-1}^2 =a_{n+1}a_{n-1}-a_n^2\implies\frac{a_{n-1}+a_{n+1}}{a_n}=\frac{a_{n-2}+a_n}{a_{n-1}}$$ This means the sequence $\big\{\frac{a_{n}+a_{n+2}}{a_{n+1}}\big\}$ is a constant. Because $a_1=a_2=1$, $a_3=\frac{1^2+2}{1}=3$, we find this constant equals $\frac{1+3}{1}=4$. This means $$\frac{a_{n}+a_{n+2}}{a_{n+1}}=4\implies a_{n+2}=4a_{n+1}-a_n$$ Because initial terms are integers, all the following terms are indeed integers.