Analogy between $\Bbb F[x]$ and $\Bbb Z$?

Yes, you can always specialize such polynomial equations via evaluation hom's

$$\begin{align} f(x) &= q(x) g(x) + r(x)\ \Rightarrow\ f(x) = r(x) \ \ {\rm in}\ \ \Bbb F[x]/q(x)\\ \Rightarrow\ \ f(a) &= q(a) g(a) + r(a) \ \,\Rightarrow\ f(a) = r(a) \ \ {\rm in}\ \ \Bbb Z/q(a) \end{align}$$

For example, casting out nines can be viewed that way when we interpret decimal notation in polynomial form, i.e $\, n = f(10) = f_0 + f_1\cdot 10+\cdots +f_n\cdot 10^k.\ $ Then $$\begin{align} f(x) &= (x\!-\!1) g(x) + f(1)\ \Rightarrow\ \,\ f(x) = f(1) \ \ {\rm in}\ \ \Bbb Z[x]/(x\!-\!1)\\ \Rightarrow\ \ f(10) &= \ \quad\ 9\ g(10) + f(1) \ \Rightarrow\, f(10) = f(1) \ \ {\rm in}\ \ \Bbb Z/9 \end{align}$$

where $\, f(1) = $ sum of digits of $\,n.\,$ For another example, let $\,f_n = (x^n-1)/(x-1),\,$ which satisfies $\, \gcd(f_m,f_n)\, =\, f_{\,\gcd(m,n)}.\,$ Specializing the associated Bezout gcd identity for $\,\gcd(m,n)=\gcd(15,21) = 3,\,$ then evaluating the polynomials at $\,x=2\,$ yields

$$\rm \begin{align} \frac{x^3-1}{x-1} &=\ (x^{15} + x^9 + 1)\ \frac{x^{15}-1}{x-1}\ -\ (x^9+x^3)\ \frac{x^{21}-1}{x-1}\\[0.5em] \Rightarrow\ \ 2^3-1\ &=\ (2^{15}+2^9+1)\ (2^{15}-1)\ -\ (2^9+2^3)\ (2^{21}-1) \end{align}$$

See this answer for more on this divisibility sequence.