The most general way I can formulate my question is the following: 


Question 1: *Given a Gorenstein quotient ring $S$ of a polynomial ring over a field $K$, can one construct a (topological) space $X$ such that the (even degree part of the) singular cohomology ring of $X$ with coefficients in $K$ is isomorphic to $S$?*


Edit: as mentioned in the comments, the answer depends on the grading of the variables. I would be most interested in a uniform grading (i.e all variables have the same degree d, without restriction on the value of d).


For the specific cases I have in mind, $S$ is a quotient of a polynomial ring over $\mathbb{C}$ by an ideal generated by monomials and binomials. Below is an example of such a ring $S$ that I would like to realize as the cohomology ring of some space:


$$ S = \mathbb{C}[x\_1,\ldots,x\_7]/(x\_7^2, x\_3x\_7-x\_4x\_7, x\_2x\_7-x\_5x\_7, x\_1x\_7-x\_6x\_7, x\_6^2, x\_3x\_6-x\_5x\_6, x\_2x\_6-x\_4x\_6, x\_1x\_6-x\_6x\_7, x\_5^2, x\_3x\_5-x\_5x\_6, x\_2x\_5-x\_5x\_7, x\_1x\_5-x\_4x\_5, x\_4^2, x\_3x\_4-x\_4x\_7, x\_2x\_4-x\_4x\_6, x\_1x\_4-x\_4x\_5, x\_3^2, x\_1x\_3-x\_2x\_3, x\_2^2, x\_1x\_2-x\_2x\_3, x\_1^2).$$


The ring $S$ above is Gorenstein and all rings that I am considering for my question are also Gorenstein (i.e satisfy Poincare duality). 


Edit 2: I will put this here rather than leaving the additional question in the comments because it goes deeper into what I was really interested in.


Question 2: *Since the answer to Question 1 seems to be affirmative, can the space X be chosen to be "nice" (e.g. compact manifold) under the assumption that $S$ (the cohomology ring of $X$) satisfies Poincare duality?*



Constructing a space with prescribed cohomology ring
I did not understand exactly your question. What is important here, is the ring $k$. If $k=\mathbf{Z}$ then the problem you are asking for is hard, if $k$ is any commutative ring then the problem is very hard. If $k$ is a field of characteristic 0 or $\mathbf{F}\_{p}$ then there is always a solution. When the characteristic is 0, the problem is solved by Sullivan (under finiteness condition + simply connected+ condition on generalized Steenrod 0-operation cf comments). I will try to explain the situation over $\mathbf{F}\_{p}$. Detnote by $\mathsf{E}\_{\overline{\mathbf{F}}\_{p}}$ the model category of $E\_{\infty}$- differential graded $\overline{\mathbf{F}}\_{p}$-algebras (in positive degree). Mandell's theorem says the following Quillen adjunction :
$$C^{\ast}(-,\overline{\mathbf{F}}\_{p}):\mathsf{sSet}^{op} \longrightarrow \mathsf{E}\_{\overline{\mathbf{F}}\_{p}}: Map\_{\mathsf{E}\_{\overline{\mathbf{F}}\_{p}}}(-,\overline{\mathbf{F}}\_{p})$$ 
which induces an equivalence between $\infty$-subcategories (I will not go to the details here). Suppose $S$ is graded commutative $\mathbf{F}\_{p}$-algebra, such that in each degree $S\_{n}$ is finite dimensional $\mathbf{F}\_{p}$-module, and $S\_{0}=\mathbf{F}\_{p}$ and $S\_{1}=0$. In particular $S\otimes\overline{\mathbf{F}}\_{p} $ is an object of $\mathsf{E}\_{\overline{\mathbf{F}}\_{p}}$.
The space you are looking for is given by the derived $\mathbb{R}Map\_{\mathsf{E}\_{\overline{\mathbf{F}}\_{p}}}(S\otimes\overline{\mathbf{F}}\_{p} ,\overline{\mathbf{F}}\_{p})\simeq \mathbb{R}Map\_{\mathsf{E}\_{\mathbf{F}\_{p}}}(S ,\overline{\mathbf{F}}\_{p}):=X$. Applying Mendell's Theorem, we obtain that 
$$C^{\ast}(X,\overline{\mathbf{F}}\_{p})\simeq S\otimes\overline{\mathbf{F}}\_{p}. $$
By definition, the cohomology of $X$ with coefficients in $\overline{\mathbf{F}}\_{p}$ is $S\otimes \overline{\mathbf{F}}\_{p}$ since $S$ is formal by definition. Then you conclude for $\mathbf{F}\_{p}$ coefficients. 


PS: the functor $C^{\ast}(-,k)$ is the functor of cochain complex. 


