The question is, Given MCQ, Which of the following is true?
(a) $Z[x]$ is principal ideal domain.
(b) $Z[x,y]/\langle y+1\rangle$ is a unique factorization domain.
(c) If $R$ is a principal ideal domain and $p$ is a non-zero prime ideal, then $R/P$ has finitely many prime ideal.
(d) If $R$ is principal ideal domain, then any subring of $R$ containing $I$ is again a principle ideal domain.
The correct option are (b) and (c). I got the option (c) is correct. For option (b), it was written in the explanation, that $\frac{\mathbb{Z[x,y]}}{\langle y+1\rangle}\cong \mathbb{Z[x]}$ and since $\mathbb{Z[x]}$ is Unique Factorization Domain, $\frac{\mathbb{Z[x,y]}}{\langle y+1\rangle}$ is also unique factorization domain.
My question, how we got that $\frac{\mathbb{Z[x,y]}}{\langle y+1\rangle}\cong \mathbb{Z[x]}$? How just seeing this quotient ring, we get the idea that this quotient ring can be isomorphic to $\mathbb{Z[x]}$?
Given $\alpha\in \mathbb{Z}[x,y]$ we can regard it as a polynomial in $y$ with coefficients in $\mathbb{Z}[x]$: $$\alpha=\sum_{i=0}^n r_i(x)y^i.$$
Then \begin{eqnarray*}\alpha&=&\sum_{i=0}^n r_i(x)(y^i-(-1)^i)+\sum_{i=0}^n (-1)^ir_i(x)\\\\&=&(y+1)\sum_{i=0}^n r_i(x)(y^{i-1}-y^{i-2}+\cdots -(-1)^i)+\sum_{i=0}^n (-1)^ir_i(x)\end{eqnarray*}
Thus in $\mathbb{Z}[x,y]/\langle y+1 \rangle$, $\alpha$ is equivalent to $\sum_{i=0}^n (-1)^ir_i(x)\in \mathbb{Z}[x]$.
Further $1+y$ does not divide any non-zero element of the subring $\mathbb{Z}[x]$, so distinct elements of this subring remain distinct in the quotient.
The above calculation is just capturing the intuition mentioned in comments.