There are indeed very many rings in which the nilradical equals the Jacobson radical.
Consider the following property of a ring: every prime ideal is the intersection of the maximal ideals containing it. These rings are called Hilbert-Jacobson rings. In a Hilbert-Jacobson ring, the intersection of all the prime ideals is therefore also the intersection of all maximal ideals, i.e., the nilradical and Jacobson radical coincide.
And indeed there are very many Hilbert-Jacobson rings. One useful result in this direction is:
Theorem: Let $R$ be a Hilbert Jacobson ring, and let $S$ be a ring which is finitely generated as an $R$-algebra. Then $S$ is also a Hilbert-Jacobson ring.
Since any field is trivially a Hilbert-Jacobson ring, it follows that all algebras which are finitely generated over a field are Hilbert-Jacobson, as Arturo Magidin pointed out in his answer.
For some information on this subject, including a proof of the theorem, see these notes.
Hint $\rm\ f\in J(R[x])\Rightarrow f\:$ in all max $\rm M$ $\Rightarrow\:\overbrace{\rm 1\! +\! x\:\!f\:\text{ in no max}\ \rm M}^{\small\textstyle \rm 1\!+\!x\:\!f,f\in M\Rightarrow 1\in M}\Rightarrow \rm 1\!+\!x\:\!f\,$ unit $\rm \Rightarrow f = 0$
Remark $ $ More generally this shows that $\rm R[x]$ has Jacobson radical equal to its nilradical: the above shows $\rm\ f\in J(R[x])\Rightarrow 1+x\:\!f\,$ a unit, so by here all coef's of $\rm\,f\,$ are in $\rm\sqrt R$ so $\rm\,f\in \sqrt{R[x]}$
Perhaps the following is of interest, from my post giving a constructive generalization of Euclid's proof of infinitely many primes (for any ring with fewer units than elements).
Theorem $ $ TFAE in ring $\rm\:R\:$ with units $\rm\:U,\:$ ideal $\rm\:J,\:$ and Jacobson radical $\rm\:Jac(R)$
$\rm(1)\quad J \subseteq Jac(R),\quad $ i.e. $\rm\:J\:$ lies in every max ideal $\rm\:M\:$ of $\rm\:R$
$\rm(2)\quad 1+J \subseteq U,\quad\ $ i.e. $\rm\: 1 + j\:$ is a unit for every $\rm\: j \in J$
$\rm(3)\quad I\neq 1\ \Rightarrow\ I+J \neq 1,\qquad $ i.e. proper ideals survive in $\rm\:R/J$
$\rm(4)\quad M\:$ max $\rm\:\Rightarrow M+J \ne 1,\quad $ i.e. $ $ max $ $ ideals $ $ survive $ $ in $\rm\:R/J$
Proof $\: $ (sketch) $\ $ With $\rm\:i \in I,\ j \in J,\:$ and max ideal $\rm\:M,$
$\rm(1\Rightarrow 2)\quad j \in all\ M\ \Rightarrow\ 1+j \in no\ M\ \Rightarrow\ 1+j\:$ unit.
$\rm(2\Rightarrow 3)\quad i+j = 1\ \Rightarrow\ 1-j = i\:$ unit $\rm\:\Rightarrow I = 1\:.$
$\rm(3\Rightarrow 4)\ \:$ Let $\rm\:I = M\:$ max.
$\rm(4\Rightarrow 1)\quad M+J \ne 1 \Rightarrow\ J \subseteq M\:$ by $\rm\:M\:$ max.
Best Answer
First note that $S\cong\bigoplus_{n\in\Bbb N}R$ as $R$-module. Moreover the jacobson radical preserves direct sums, hence $$J_R(S)\cong\bigoplus_{n\in\Bbb N}J_R(R)$$ that's the submodule of polynomials with coefficients in $J_R(R)$.
To prove that the Jacobson radical commutes with direct sum of modules, first note that every $R$-module homomorphism $\varphi:M\to N$ maps $J_R(M)$ into $J_R(N)$. Applying this to the canonical projections $\bigoplus_iM_i\to M_i$ gives $J_R(\bigoplus_iM_i)\subseteq\bigoplus_iJ_R(M_i)$. Similarly, by considering the canonical inclusions $M_i\to\bigoplus_iM_i$ we get the reverse inclusion $J_R(\bigoplus_iM_i)\supseteq\bigoplus_iJ_R(M_i)$.