Solve for all $x$ such that $x^3 = 2x + 1, x^4 = 3x + 2, x^5 = 5x + 3, x^6 = 8x +5 \cdots$ Question:
Solve for all $x$ such that $\begin{cases}&{x}^{3}=2x+1\\&{x}^{4}=3x+2\\&{x}^{5}=5x+3\\&{x}^{6}=8x+5\\&\vdots\end{cases}$.

My attempt:
I sum up everything.
$$\begin{aligned}\sum_{i=1}^n x^{i+2} &= (2 + 3 + 5 + 8 + \cdots)x + (1 + 2 + 3 + 5 +...)\\&= (S_n - 2)x + (S_n - 1)\\& = (a_{n+2} - 3)x + (a_{n+2} - 2)\end{aligned}$$
where $S_n$ is sum of first $n$ terms of Fibonacci series and $a_n$ is its $n$th term.
I'm not getting any idea how to solve it further.

$x^3 = 2x + 1$  has an evident rational solution, that being $-1.$  Next, $x+1$ divides $x^3 - 2x  - 1,$  the quotient is $x^2 - x-1$  for
$$  x^3 - 2x - 1 = (x+1) (x^2 - x - 1)  $$
However $-1$  does not satisfy $x^4 = 3x + 2.$
We are left with those numbers $x$ with $x^2 = x+1.$  We can find the values of $x^n$  by a simple rule:   if $x^n = ax + b,$  then
$$x^{n+1} = a x^2 + bx = a (x+1) + bx = (a+b)x + a.  $$
The coefficient pairs  for $x^n$   come out
$$ 
\begin{array}{ccc}
2: & 1 & 1 \\
3: & 2 & 1 \\
4 : & 3 & 2 \\
5 : & 5 & 3 \\
\end{array}
$$
and so on. Given a row $(a,b) $ the next row  is what happens when we multiply it on the right side by
$$ M= 
\left(
\begin{array}{cc}
 1 & 1 \\
 1 & 0 \\
\end{array}
\right)
$$
All of the pairs are of the form $(1 \hspace{2mm} 1) \; M^j . $  Actually, now that I think of it, all of the pairs are of the form  $(1 \hspace{2mm} 0) \; M^k  $  and are just the top row of that $M^k. \;$ These are, of course, consecutive fibonacci numbers
