Find function with a degree of 4 from three statements If $f(x)$ is a function/polynomial with a degree of 4 and:
$$f(-1)=1$$
$$f(-2)=4$$
$$f(-3)=9$$
How can I find $f(x)$?

From statement 1, I know $f(x) = (x+1)(...)+1$
With statement 2, I know $f(x) = (x+1)[(x+2)(...)-3]+1$
We can use a similar way to find $f(x)$ with statement 3, but don't we end up with a cubic polynomial?
If my calculations are correct, how can we turn the cubic into a 4 degree polynomial?
This is a plain application of polynomial interpolation.
Three statements make it possible to uniquely define a polynomial of degree $2$. You can still find all the polynomial of degree $n\geq 2$ that satisfy those statements (using the Lagrange basis for example).
In this answer, I will determine the necessary and sufficient condition for the polynomial degree to be at least $4.$
I considered your polynomial to be a quartic of at least degree $4.$ But, this is not enough.
Method
For the quartic polynomial, at least value of $f(-4)$ must be known. Let, $f(-4)=p$, then you have
$$f(x)=a(x+1)(x+2)(x+3)+x^2$$
$$\implies a×(-6)+16=p$$
$$\implies a=\dfrac {16-p}{6}$$
$$\implies f(x)=\left(\dfrac {16-p}{6}\right)(x+1)(x+2)(x+3)+x^2$$
But, this is still not enough. Because, we have a cubic polynomial. You still need value of $f(-5).$
Let,
$$f(-5)=\left(\dfrac {16-p}{6}\right)\times (-24)+25\Longrightarrow f(-5)=4p-39$$
So, if $f(-5)=4f(-4)-39$ you still have a cubic polynomial.
Let, $f(-4)=p$ and $f(-5)≠4f(-4)-39$, then our tricky method will not work. So, you need a general method.
$$\begin{cases} a-b+c-d+e=1 \\16a-8b+4c-2d+e=4\\ 81a-27b+9c-3d+e=9 \\ 256a-64b+16c-4d+e=f(-4)\\ 625a-125b+25c-5d+e=f(-5) \end{cases}$$
You can solve this linear system with Gauss method.
A necessary and sufficient condition:
At least the values of  ​​$f(-4)$ and $f(-5)$ must be given and the following condition must be hold:
$$f(-5)-4f(-4)+39≠0.$$
Conclusion
To obtain a polynomial of degree of at least $4$, it is necessary to know at least the values ​​of $f(-4)$ and $f (-5)$, where $f(-5)≠4f(-4)-39$ or $f(-5)-4f(-4)+39≠0.$
To adapt to your circumstances, I wrote $f (-4)$ and $f(-5)$ instead of $f(4)$ and $f(5).$
Let's test the necessary and sufficient condition.
Let,  $$ f(-1)=1 \\ f(-2)=4 \\ f(-3)=9\\ f(-4)=16 \\ f(-5)=25 $$
Then, we have,
$$f(5)-4f(-4)+39 \Longrightarrow 25-4 \times 16+39=0$$
As it seems, the necessary and sufficient condition are not provided.  So, we can find a lower degree polynomial. And we really have such a function:
For example, the function $f(x)=x^2$ is the simplest function that provides the values ​​we give.
Remark
Here we have only examined the requirement to give at least $f(-4)$ and $f(-5)$ values.  However, it is also possible to write it for the cases $f(4)$ and $f(5)$, or $f(7)$ and $f(9)$. However, this can never invalidate the condition $f(-5)-4f(-4)+39≠0.$