Let $k$ be a field and $k[x]$ be the polynomial ring. Let $R$ be a commutative ring with unity and further assume that it is not a field. Let $m$ a maximal ideal of $R$ such that $R/m \simeq k$.
Then does there always exist a ring morphism $\phi : R \rightarrow k[x]$ such that ideal $(x)$ in $k[x]$ contracts to $m$ , that is, $\phi^{-1}(x)=m$. If that is true, then is it also true that given two maximal ideals $m_{1}$ and $m_{2}$ (such that $R/m_{i} \simeq k $ for $i=1,2$ ), there exists a morphism $\phi : R \rightarrow k[x]$ such that ideal $(x)$ contracts to $m_{1}$ and $(x-1)$ contracts to $m_{2}$.
In the first case, yes: just let $R$ be the composition of the quotient map $R\to R/m\cong k$ with the inclusion of the constants $k\to k[x]$.
In the second case, no. For instance, let $k=\mathbb{Q}$, $R=\mathbb{Q}[x,x^{-1}]$, $m_1=(x-1)$ and $m_2=(x-2)$. Note that the image of any homomorphism $\phi:R\to\mathbb{Q}[x]$ consists entirely of constants, since $x$ must map to a unit and the only units in $\mathbb{Q}[x]$ are constants. So for any such $\phi$, $\phi^{-1}(x)=\phi^{-1}(x-1)=\ker(\phi)$.
Geometrically, if $R$ is a $k$-algebra and you ask for $\phi$ to be a homomorphism of $k$-algebras, you are asking whether any two $k$-points of $\operatorname{Spec} R$ can be connected by a morphism from $\mathbb{A}^1$. This is usually false.