I seem to have arrived at a contradiction by applying what I know about quotient rings. I can't figure out where the mistake is.
Let $f(x)\in \mathbb{Z}[x]$, $\deg f(x)\ge 1$, and $p$ a prime number. Then
\begin{align*} \mathbb{Z}[x]/f(x)\mathbb{Z}[x]&\cong (\mathbb{Z}[x]/p\mathbb{Z}[x])/((f(x)\mathbb{Z}[x])/p\mathbb{Z}[x])\tag{Third Iso. Thm}\\ &\cong(\mathbb{Z}[x]/p\mathbb{Z}[x])/(\hat{f}(x)(\mathbb{Z}[x]/p\mathbb{Z}[x]))\tag{$*$}\\ &\cong \mathbb{Z}_p[x]/\hat{f}(x)\mathbb{Z}_p[x]. \end{align*}
where $(*)$ comes from:
\begin{align*} (f(x)\mathbb{Z}[x])/p\mathbb{Z}[x]&=\{g(x)+p\mathbb{Z}[x]:g(x)\in f(x)\mathbb{Z}[x]\}\\ &=\{f(x)g(x)+p\mathbb{Z}[x]:g(x)\in\mathbb{Z}[x]\}\\ &=\{(f(x)+p\mathbb{Z}[x])(g(x)+p\mathbb{Z}[x]):g(x)\in\mathbb{Z}[x]\}\\ &=\{\hat{f}(x)(g(x)+p\mathbb{Z}[x]):g(x)\in\mathbb{Z}[x]\}\\ &=\hat{f}(x)(\mathbb{Z}[x]/p\mathbb{Z}[x]) \end{align*}
For example, this would seem to show that $$\mathbb{Z}[i]\cong \mathbb{Z}[x]/(x^2+1)\mathbb{Z}[x]\cong \mathbb{Z}_2[x]/(x^2+1)\mathbb{Z}_2[x]$$ The first ring is infinite and the last ring is finite. What's going on here?
Whatever you meant, the very first equality (isomorphism) is wrong unless $\;p\Bbb Z[x]⊂f(x)\Bbb Z[x]\;$ , as required by the third isomorphism theorem