Is $\operatorname{Spec}(\mathbb{C}[x,y]/(y-x^2))$ the same as $\operatorname{Spec}(\mathbb{C}[x])$ ? Let $R=\mathbb{C}[x,y]/(y-x^2)$. We have that $$\operatorname{Spec}(R)=\{(0),(x-a,y-a^2),(y-x^2)\}. $$
But if we consider the quotient ring $\mathbb{C}[x,y]/(y-x^2) \simeq \mathbb{C}[x]$. But $\operatorname{Spec}(\mathbb{C}[x])=\{(0),(x-a)\}$. So we have isomorphic rings that have different $\operatorname{Spec}$... How can I solve this problem?
$(x-a,y-a^2)\equiv(x-a)$ since $y-a^2\equiv x^2-a^2\equiv(x-a)(x+a)$ in the quotient. 

Deleting the $(0)$ (which looks rather like a typo) from the first list, your two lists match.
There is no problem: the element $a$ in your description of $\mathrm{Spec}(R)$ can be any complex number and different $a$'s lead to different prime ideals. Hence the spectrum is infinite, as well as the spectrum of the polynomial ring.