Observe the isomorphism $$ \mathbb{Z}[x]/(x-2,x+3)\cong\big(\mathbb{Z}[x]/(x-2)\big)/(x+3). $$ Since $x+3$ is irreducible modulo $x-2$, the RHS of the above isomorphism is an integral domain (since irreducible polynomial generates prime ideal). Thus the LHS is an integral domain, so $(x-2,x+3)$ is a prime ideal. Is this right?
For the part about not being principal, assume $(x-2,x+3)=(p(x))$ for some $p(x)\in\mathbb{Z}[x]$. This means $p(x)$ is a common divisor of both $x-2$ and $x+3$. But $\gcd(x-2,x+3)=1$, so $p(x)=1$. However, this means $(p(x))=\mathbb{Z}[x]$, and since $1\not\in(x-2,x+3)$, we know $(x-2,x+3)\not=\mathbb{Z}[x]$. Thus the ideal cannot be generated by a single polynomial. Is this correct?
The quotient ring is indeed a field (it's isomorphic to $\mathbb{F}_{5}$). If you take the quotient with a principal ideal in this ring you can never obtain a field.