Here is a perhaps slightly more concrete way of looking at things. In the longer run, the more abstract approaches are essential.
What are the elements of $\mathbb{Z}[x]/(x^2+1)$? They are equivalence classes. Two polynomials are called equivalent if they differ by a multiple of the polynomial $x^2+1$.
In our situation, we can give a more concrete description of the equivalence classes. Let $P(x)$ be a polynomial with integer coefficients. Imagine dividing $P(x)$ by $x^2+1$, using the ordinary procedure for dividing polynomials. We get in general a remainder which is a polynomial of the shape $bx+a$ (one or both of $a$ and $b$ may be $0$).
So we have $P(x)=Q(x)(x^2+1) +bx+a$ where $Q(x)$ is the quotient. If you look at the above displayed formula, you will see that $P(x)$ and $bx+a$ are equivalent. Formally, $P(x) +(x^2+1)= bx+a + (x^2+1)$ It is easy to see that if $bx+a$ and $dx+c$ are different polynomials, then they determine different equivalence classes, since their difference is not divisible by $x^2+1$. Thus every equivalence class corresponds to a unique polynomial of the form $bx+a$.
This is analogous to what happens with congruence modulo $m$. The collection of equivalence classes there is $\mathbb{Z}/(m)$, but we can think of $\mathbb{Z}/(m)$ as (sort of) $0, 1, 2, \dots, m-1$ with a funny addition and multiplication.
Now where should we send $a+bx+(x^2+1)$? By basic properties of isomorphism, once we know where $1+(x^2+1)$ and $x+(x^2+1)$ are to go, the rest is determined. It is natural, indeed necessary, to send the equivalence class of $1$ to $a+bi$, where $a=1$ and $b=0$.
Where should $x+(x^2+1)$ go? It is reasonable to send it to $i$. So then the general object $a+bx +(x^2+1)$ should go to $a+bi$.
Does this work? You need to show this mapping is one-to-one and onto, or to use more currently approved language, that the mapping is a bijection. That is easy, given the prior discussion.
Then we need to show that the mapping "preserves" addition and multiplication. The fact that addition is preserved is mechanical. Multiplication is a bit trickier. The product $[P(x)+(x^2+1)][Q(x)+(x^2+1)]$ is, by definition, $P(x)Q(x)+(x^2+1)$.
Apply this to $P(x)=a+bx$, $Q(x)=c+dx$. Note that $(a+bx)(c+dx)=ac +bdx^2 +(ad+bc)x$.
But $bdx^2=bd(x^2+1)-bd$, so $bdx^2$ is equivalent to $-bd$. Putting things together, we find that $(a+bx)(c+dx)$ is equivalent to $ac-bd +(ad+bc)x$. This should remind you of the rule for multiplying complex numbers, and help to take care of showing that the mapping behaves well under multiplication.
A bit of intuition The polynomial $x^2+1$ is equivalent to the $0$-polynomial, since they differ by a multiple of $x^2+1$. So in $\mathbb{Z}[x]/(x^2+1)$, "$x^2+1$" behaves like $0$, so "$x^2$" behaves like "$-1$". Thus the square of "$x$" is "$-1$".
More formally, $[x+(x^2+1)]^2 =x^2+(x^2+1) =-1+(x^2+1)$
Note: There is another mapping that would give an isomorphism, namely the one that takes $a+bx +(x^2+1)$ to $a-bi$.