Let $\mathbb{N}$ denote the set of positive integers. Does there exist a function $f:\mathbb{N}\rightarrow\mathbb{N}$ such that $$f(f(n))=f(n-1)f(n+1)-f(n)^2$$ for all $n\geq 2$?
If $f$ is linear, plugging in $f(n)=an+b$, we get $$a(an+b)+b=(a(n-1)+b)(a(n+1)+b)-(an+b)^2$$ which is $$a^2n+ab+b=-a^2$$
This means $a=0$. But then also $b=0$, hence $f\equiv 0$, but this is not allowed. So there is no such linear function.
Since the values of $f$ must be positive, $f(n-1)f(n+1) > f(n)^2$ for all $n \geq 2$, so $\frac{f(n+1)}{f(n)} > \frac{f(n)}{f(n-1)}$.
Therefore there exist $M, R > 1$ such that for all $n > M$, $f$ is increasing and $f(n) > nR^n > n + 2$. For such $n$,
$$f(f(n)) < f(n-1)f(n+1) < f(n+1)^2$$
On the other hand, $$f(f(n)) > f(n + 2) > (f(n+2)f(n))^{1/2} > (f(f(n+1)))^{1/2}$$ $$> f(n+1)^{1/2}R^{f(n+1)/2}$$
Therefore $$f(n+1)^{1/2}R^{f(n+1)/2} < f(n+1)^2$$ $$R^{f(n+1)/3} < f(n+1)$$
But since $R > 1$ and $f$ is increasing for $n > M$, this cannot hold for all such $n$. Therefore no such $f$ exists.