From [This short paper](https://www.researchgate.net/publication/382801850_Conjectured_Somos-like_closed_form_of_recurrences_with_polynomial_coefficients)

For polynomial $F$ with integer coefficients, define the recurrence
$f(n)=F(n,f(n-1),f(n-2),...,f(n-d))$. We conjecture that
$f(n)$ satisfy Somos like sequence  
$f(n)=\frac{G(f(n-1),...,f(n-d_1))}{H(f(n-1),...,f(n-d_2))}$
for polynomials $G$,$H$ which do not depend on $n$.

The strong conjecture is for all $F$ and the weak conjecture
is about $F$ which is linear in $f(n-i)$.

It solves special parametric cases like
$f(n)=F(n,f(n-1),f(n-2))=(f(n-1)^2)(b_1 n+b_2)+f(n-2)(b_3 n+b_4)$ for
all integers $b_i$ and $f(n)=f(n-1)(n^2+n+2)+f(n-2)(3 n+ 5 )$.
The proofs are based on finding algebraic dependency of
polynomials.

>Q1 Are the conjectures true? 

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Added later

In comments **Max Alekseyev** proves existence of algebraic dependency
not depending on $n$ using resultants of consecutive terms.

This is not enough for a proof, because in addition we need the
dependency to be linear in the leading term.

The paper discusses this issue and in this case of failure
it recommends to increase the "order" of the algebraic dependency.

In a deleted answer Max suggests the counterexample
$f(n) = (n+1)^2 f(n-1) + n^2 f(n-2)$, but our sage implementation
found closed form of order $5$ using algebraic dependency, 
disproving the counterexample.