Proving that a prime ideal $p \subset R$ yields a prime ideal $p[x] \subset R[x]$ I'm curious as to whether I can have my proof critiqued. 

Proposition : Let $\mathfrak{p}$ be a prime ideal in a ring $R$. Show that $\mathfrak{p}[x]$ is a prime ideal in $R[x]$. 

Proof : Suppose $\mathfrak{p}$ is  a prime ideal in a ring $R$. Then $\mathfrak{p} \subset R$. Therefore, consider that $$R \otimes_R \bigoplus_{i=1}^{\infty} R \cong R[x].$$ Since $\mathfrak{p} \subset R$, we have that $$\mathfrak{p} \otimes_R \bigoplus_{i=1}^{\infty} R \cong \bigoplus_{i=1}^{\infty} \mathfrak{p} \otimes_R R \cong \mathfrak{p}[x] \subset R[x].$$ Given that prime ideal are preserved under any isomorphism, it follows that $\mathfrak{p}[x]$ is a prime ideal in $R[x]$.

I furthermore claim that since isomorphisms do not necessarily preserve maximal ideals, if $\mathfrak{m} \subset R$ is a maximal ideal in $R$, it is not necessarily true that $\mathfrak{m}[x] \subset R[x]$ is a maximal ideal in $R[x]$.
It is infact never true that $\mathfrak{m}[x]\subseteq R[x]$! All of these questions become simple when you consider the following fact. 

An ideal $\mathfrak{p}\subseteq R$ is prime if and only if $R/\mathfrak{p}$ is an integral domain. Moreover $\mathfrak{p}$ is maximal if and only if $R/\mathfrak{p}$ is a field.

The first statement is simple, the second might require a little bit of thought. Now note that $R[x]/\mathfrak{p}[x]=R/\mathfrak{p}[x]$, a polynomial ring on an integral domain, and thus an integral domain! But $R/\mathfrak{m}[x]$ is the polynomial ring on a field, and so contains the ideal $(x)$, which proves it is nonmaximal.

Lastly, I'm not really sure about your proof, since your isomorphism is not a ring isomorphism, and appears to be a module isomorphism, which doesn't help with proving an ideal is prime as far as I know.
It is infact never true that $\mathfrak{m}[x]\subseteq R[x]$! All of these questions become simple when you consider the following fact. 

An ideal $\mathfrak{p}\subseteq R$ is prime if and only if $R/\mathfrak{p}$ is an integral domain. Moreover $\mathfrak{p}$ is maximal if and only if $R/\mathfrak{p}$ is a field.

The first statement is simple, the second might require a little bit of thought. Now note that $R[x]/\mathfrak{p}[x]=R/\mathfrak{p}[x]$, a polynomial ring on an integral domain, and thus an integral domain! But $R/\mathfrak{m}[x]$ is the polynomial ring on a field, and so contains the ideal $(x)$, which proves it is nonmaximal.

Lastly, I'm not really sure about your proof, since your isomorphism is not a ring isomorphism, and appears to be a module isomorphism, which doesn't help with proving an ideal is prime as far as I know.