OK, so let me denote the elements of End$(⟨\mathbb{Z},+⟩)$ by $\theta_{i}$ , where we have $\theta_{i}(1)=i$ for some $i \in \mathbb{Z}$.
Also I define the evaluation homomorphism as $\alpha(\theta_{i})=\theta_{i}(1)$.
Now we can see that $\alpha:$End$(⟨\mathbb{Z},+⟩)$ $\rightarrow$ $⟨\mathbb{Z},+,\cdot⟩$, and i need to show that in fact $\alpha$ is a ring isomorphism.
First ill show its a ring homomorphism, to do this, observe that$\alpha(\theta_{i} + \theta_{j})=(\theta_{i} + \theta_{j})(1)=\theta_{i}(1) + \theta_{j}(1)=\alpha(\theta_{i})+\alpha(\theta_{j})$.
And also that $\alpha(\theta_{i}\theta_{j})=\theta_{i}(\theta_{j}(1))=\theta_{i}(j)=ij=\alpha(\theta_{i})\alpha(\theta_{j})$, so it is a ring homomorphism.
Now i just need to show its one-to-one and onto.
one-to-one: if $\alpha(\theta_{i})=\alpha(\theta_{j})$, then $\theta_{i}(1)=\theta_{j}(1)$, and since the homomorphisms are determined by what values they send the generators to, we must have $\theta_{i}=\theta_{j}$, i.e they are the same homomorphism. Hence $\alpha$ is one-to-one.
Onto: now $\forall x \in \mathbb{Z}$, we have $\theta_{x}$ as defined above being the homomorphism sending $1$ to $x$, so $\alpha(\theta_{x})=x$, so we can see that $\alpha$ is obviously onto.
One invariant of a ring is its characteristic, the smallest natural number $n$ such that
$$
\underbrace{1+\ldots+1}_{\text{n times}}=0
$$
The two rings you are considering have different characteristics.
Best Answer
If $I$ is a maximal ideal in $\def\CC{\mathbb C}\CC[X]$, then there is an $\alpha\in\CC$ such that $I=(X-\alpha)$, and using this it is easy to see that $\dim_\CC I/I^2=1$.
On the other hand, the ideal $J=(X,Y)\subset A=\CC[X,Y]/(X^2-y^3)$ is maximal and $J/J^2$ is a vector space of dimension $2$.
It follows that $A$ is not isomorphic to $\CC[X]$.