How to show $\Bbb Z[x]/<3,x^2+1>$ is field? My attempt
First I wanted to show $<3,x^2+1>$ is maximal 
So, I supposed another maximal $A$ which contain $<3,x^2+1>$ properly, and choose element $a$ in $A \setminus <3,x^2+1>$ to induce $A$ must contain $1$. But this approach can be applied only to the case of $\Bbb R[x]$
Second, by failure to above approach, I want to show that  $\Bbb Z[x]/<3,x^2+1>$ is isomorphic to $Z_5[x]$ because I already know about $Z[i]/<2-i>$ is isomorphic to $Z_5[x]$. So I preferentially narrowed down given ring into the case of $Z[x]/<x^2+1>$. As result of the investigation, all elements of the ring could be expressed by $a+bi+<x^2+1>$ with assuming $x=i$ Thus I can conclude that $Z[x]/<x^2+1>$ is isomorphic to $Z_5[x]$ which is field. But I cannot connect this result with our given ring. 
I want to get some advice from your help. Since I don't learn about the polynomial ring such as association with irreducible property(?). So please give me a hint to prove that by using basic property of field.
 A: In this case, it is possible to simply compute the quotient ring. Below are some hints to guide you, but first let me make a remark.
In $\Bbb Z[x]/(3,x^2 + 1),$ the class of $3$ (i.e., $3 + (3, x^2 + 1)\in\Bbb Z[x]/(3,x^2 + 1)$) is equal to $0,$ because $$3 - 0 = 3\in (3,x^2 + 1).$$ Thus, $\Bbb Z[x]/(3,x^2 + 1)$ cannot be isomorphic to $\Bbb Z_5[x],$ as $3\neq 0$ in $\Bbb Z_5[x].$
Hint 1:

 Use the following fact: if $R$ is a ring and $a,b\in R,$ $$R/(a,b)\cong (R/(a))/(b)\cong (R/(b))/(a).$$ That is, if you have an ideal generated by multiple elements, you can quotient by those elements in any order to compute the quotient ring.

Hint 2:

 Here's another useful fact: Let $R$ be a ring and let $a\in R$ be some element. If $S$ is the quotient ring $S = R/(a),$ then $R[x]/(a)\cong S[x].$

Hint 3:

 Fact: if $k$ is a field, and $f\in k[x],$ then $k[x]/(f)$ is a field if and only if $f$ is irreducible.

Hint 4:

 One last fact: if $k$ is a field and $f\in k[x]$ has degree less than or equal to $3,$ $f$ is reducible over $k$ if and only if $f$ has a root in $k. $ That is, if $\deg f\leq 3,$ then $f$ is reducible over $k$ if and only if there exists $a\in k$ such that $f(a) = 0.$

A: Let $P(x)=\sum_{i=0}^na_ix^i\in \mathbb{Z}[x]$. By the division algorithm, we can write $$P(x)=Q(x)(x^2+1)+r(x)$$
where $Q(x),r(x)\in \mathbb{Z}[x]$ are defined uniquely such that $\deg(r)<2$. Hence $r(x)=ax+b$ for some $a,b\in \mathbb{Z}$.
Write $\overline{P(x)}$ for the the class of $P(x)$ in $\mathbb{Z}[x]/\left\langle 3,x^2+1\right\rangle$, then the above yields $\overline{P(x)}=\overline{r(x)}$. Since $3\in \left\langle 3,x^2+1\right\rangle$ any multiple of $3$ is killed as well, hence $$\overline{P(x)}=(a\mod3)\overline{x}+(b\mod 3)\overline{1}
.$$
So any element of $\mathbb{Z}[x]/\left\langle 3,x^2+1\right\rangle$ can be written as $$ax+b$$ where $a,b\in \mathbb{Z}_3$. It is clear that this quotient is a commutative ring with unit, hence we only need to show that any (non-zero) element has an inverse. You can do this explicitly: Indeed, notice that
$$(ax+b)(ax-b)=a^2x^2-b^2=-(a^2+b^2) \:\:\;\;\;\;\;\;\mbox{ in }\:\:\;\;\mathbb{Z}[x]/\left\langle 3,x^2+1\right\rangle$$
From the above equation you can easily find general inverses.
