Alternative ways of showing that $x^2 + 1$ is prime in $\mathbb{Z}[x]$.

Question

Let rings be associative, commutative, and unital in this question.

For context and motivation, I am interested in ways of showing that elements of rings are prime. Right now the main method I know involves appealing to the definition. I characterize when $a$ satisfies $p \mid a$ and then try to show:

$$ p \nmid a \;\;\text{and}\;\; p \nmid b \;\;\text{implies} \;\; p \nmid ab $$

Or, equivalently, that the complement of the ideal $(p)$ is multiplicatively closed (and $p$ is not $0$).

This method is really slow, though, and coming up with a nice way to characterize $p \mid a$ can be a lot of work.

What other ways there are to demonstrate that specific elements are prime in "easy rings" like $\mathbb{Z}[x]$ that don't require much machinery to define? I'm picking $\mathbb{Z}[x]$ as a specific, concrete example in order to keep the question relatively narrow.

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Let's call a ring element special if it is zero or a unit.

An irreducible element is a non-special element $x$ such that for all $a,b$ such that $ab = x$, $a$ is a unit or $b$ is a unit.

A prime element is a non-special element $x$ such that for all $a, b$, if $x \nmid a$ and $x \nmid b$, then $x \nmid ab$.

I'm interested in the ring $\mathbb{Z}[x]$ and the element $x^2+1$.

It is straightforward to show that $x^2+1$ is irreducible.

Let $a$ and $b$ be factors of $x^2+1$ such that $ab = x^2 + 1$.

Suppose $a$ or $b$ is zero, then $ab$ is zero which is not $x^2 + 1$, which contradicts the hypothesis.

Suppose $a$ has degree three or greater, then $ab$ would have degree three which contradicts the hypothesis.

Suppose $a$ has degree two, i.e. $a$ is of the form $ex^2+cx+d$. $b$ would have to have degree $0$, so $ab$ would equal $bex^2 + bcx + bd$. $b, d, e$ must all be units. So $c$ must be $0$. Thus, $a$ must be $1+x^2$ or $-1-x^2$.

Suppose $a$ has degree one, $b$ would have to have degree $1$ as well. Let $a$ be $cx + d$ and $b$ be $ex + f$. $ab$ would be $cex^2 + (cf+de)x + df$. $c,e,d,f$ are all units. $cf+de$ must be zero. Without loss of generality, suppose $c=f=1$. Thus $d = 1$ and $e = -1$ or vice versa. However, $(x+1)(x-1) = x^2 - 1$, which is not $x^2+1$ and thus contradicts the hypothesis.

If $a$ has degree zero, then $b$ would have degree $2$ and we would be back in the degree $2$ case.

Proving that $x^2 + 1$ is prime, however, is harder.

So far, here's what I have:

Let $a$ be an element of $\mathbb{Z}[x]$ and let $z$ be a complex number, let $a \triangleright z$ denote the evaluation of $a(x)$ where $a$ is construed as an element of $\mathbb{C}[x]$.

For example, $2x^3+4$ is in $\mathbb{Z}[x]$. It then follows that $(2x^3+4) \triangleright i = 4 - i$.

I claim that $x^2 + 1 \mid a$ if and only if $a \triangleright i = 0$.

Suppose $x^2 + 1$ divides $a$, then $a$ is expressible as $(x^2+1)b$. $i^2 + 1$ is $0$, thus $a \triangleright i$ is $0$.

Suppose $a \triangleright i = 0$. Since $a$ has real coefficients when construed as a complex polynomial, it follows that $-i$ is also a root of this polynomial, so $a \triangleright -i = 0$.

It follows that $a$ is of the form $(x^2+1)\sum_{k} c_kx^k$.

The entries is $c_k$ must all be integers. $c_0$ and $c_1$ must be integers, and the coefficient of $x^2$ is $c_0 + c_2$. $c_0$ is already an integer, so we can conclude that $c_2$ must be one as well. The same argument applies inductively for the remaining coefficients of $a$.