Le $\Bbb{K}$ be a field. I want to show wether $\frac{\Bbb{K}[X]}{\langle X^2 \rangle}$ and $\frac{\Bbb{K}[X]}{\langle X^2 + 1\rangle}$ are isomorphic.
First thing to note is that there is an nilpotent element $[X]$ in the first quotient. But in the second one I think it will depend on the roots of $X^2+1$ because for example in $\Bbb{F}_2$, $x^2+1=(X-1)^2$ so $[X-1]$ is nilpotent. So in the case where it can't be factorized there is no nilpotent element (how to prove that ?). Otherwise I don't see any morphism we can exhibit.
Any hint ?
I will change your $\Bbb K$ to $K$, for simplicity. All isomorphism below are isomorphisms of rings (or $K$-algebras, actually).
The structure of $K[X]/(X^2 + 1)$ depends on the number of different roots of the polynomial $X^2 + 1$ in $K$. There can be three situations:
There are two different roots $r_1, r_2$. We then have $$K[X]/(X^2 + 1) \simeq K[X]/(X - r_1) \times K[X]/(X - r_2)$$ by chinese remainder theorem, and each factor $K[X]/(X - r_i)$ is canonically isomorphic to $K$. Hence $K[X]/(X^2 + 1) \simeq K^2$ in this case.
There is one double root $r$. It is easy to see that this happens iff the characteristic of $K$ is equal to $2$, and in that case we have $X^2 + 1 = (X + 1)^2$, hence $$K[X]/(X^2 + 1) \simeq K[X]/(X + 1)^2 \simeq K[X]/X^2$$ via evident isomorphisms.
There is no root, i.e. $-1$ is not a square in $K$. In this case, $K[X]/(X^2 + 1)$ is a field, which is the quadratic extension $K[\sqrt{-1}]$ of $K$.
We see that in the second case, we do have $K[X]/(X^2 + 1) \simeq K[X]/X^2$.
In the remaining two cases, $K[X]/(X^2 + 1)$ is a product of fields, hence does not have any nilpotent element, while $X$ is a nilpotent element in $K[X]/X^2$. Thus they are not isomorphic in these two cases.
To summerize, we have $K[X]/(X^2 + 1) \simeq K[X]/X^2$ iff the characteristic of $K$ is equal to $2$.