This is a probably known consequence of the von Staudt-Clausen theorem.
From the exponential generating function
$$G(x,t)~=~ \sum S_p(x) \frac{t^p}{p!} ~=~ e^t\,\frac{1-e^{x\,t}}{1-e^t}$$
of the polynomials $S_p(x)$ it follows that a rational zero $x_0=m/n$ with $n\geq 1$ and $\gcd (n,m)=1$ leads to a zero coefficient of the series $G(x_0,t)$, and accordingly of
$$-G(\frac{m}{n},-n\,t) ~=~ \frac{1-e^{-m\,t}}{1-e^{ n\,t}}. $$
Evidently there are zero coefficients if $m=0$, $(m,n)=(-1,1),(-1,2)$, corresponding to the factors $x$, $x+1$, and $2\,x +1$, whereby the latter arise from the vanishing nonlinear odd terms in the expansion of $$f(t)~=~\frac{1 }{1+e^{ t}} ~~{\rm due~to}~~f'(t)~=~-f'(-t). $$
From the exponential generating function for the Bernoulli numbers $B_k=B_k^-$ one has
$$ - \frac{n\,t}{1-e^{ n\,t}} ~=~ \sum \limits_{k=0}^\infty \frac{B_k \,n^k\,t^k}{k!}~=~T\,e^{n\,b\,t}, $$
where $T$ is the linear umbral operator with $T b^k ~:=~ B_k$.
$$n\,G(\frac{-m}{n},-n\,t) ~=~T\, \frac{e^{(m+n\,b)\,t}-e^{n\,b\,t}}{ t}\,~=~T\, \sum \limits_{k=1}^\infty \left( (m+n\,b)^k -(n\,b)^k\right)\frac{t^{k-1}}{k!}. $$
This gives the condition for a rational root $-m/n$, that there is a $k>1$,such that
$$ T\,(m+n\,b)^k ~=~ \sum \limits_{\ell =0 }^k {k \choose \ell} m^{k-\ell}\,n^\ell\,B_\ell ~=~ n ^k B_k,$$
or equivalently for $k>2$
$$0~=~ \sum \limits_{\ell =0 }^{k-1} {k \choose \ell} \left(\frac{m}{n}\right)^{k-\ell} \,B_\ell ~=~ \left(\frac{m}{n}\right)^k-\frac{k}{2}\,\left(\frac{m}{n}\right)^{k-1} \,+ \sum \limits_{\ell =1 }^{\lfloor \frac{k-1}{2} \rfloor} {k \choose 2\,\ell} \left(\frac{m}{n}\right)^{k-2\ell} \,B_{2\,\ell} .$$
From the von Staudt-Clausen theorem it follows that the denominator of $B_{2\,\ell}$ is square free and equals the product of all primes $p$ for which $p-1 | 2\ell$, such that for the $p$-adic norm $|B_{2\,\ell}|_p\leq p$, and either $|B_{2\,\ell}|_p=1$ or $|B_{2\,\ell-2}|_p=1$ for $p>3$. Because the first polynomials $S_k(x)$ have no rational roots besides 0,-1,-1/2 it is also assumed that e.g. $k>5$.
Now for primes $p\geq2$ the following cases cover the remaining possibilities for $(n,m)=1$: (1) $n>2$ ~(2) $n\leq2$ and $m>n$,~(3) $m=-1,~n=2$.
Case 1:
In this case there is a prime $p>2$ with $v_p(n)=s\geq 1$, or $p=2$ and $s>1$. For the terms
$$A_\ell ~=~{k \choose \ell} \left(\frac{m}{n}\right)^{k-\ell} \,B_\ell $$
in the right sum one then has for $\ell>0$:
$$\left|A_0 \right|_p ~=~ p^{s\,k},
~~~\left|A_{2\ell} \right|_p \leq p^{s(k-2\ell)+1},
~~~\left|A_{2\ell} \right|_p~\leq~\left|A_1 \right|_p <\left|A_0 \right|_p.$$
It follows that
$$~\left| \sum \limits_{\ell =1 }^{k-1} A_\ell \right|_p ~\leq~\left|A_1 \right|_p <\left|A_0 \right|_p,$$
and therefore
$$~\left| \sum \limits_{\ell =0 }^{k-1} A_\ell \right|_p ~=~ p^{sk} \neq 0.$$
It is thus impossible that $n$ is any integer other than 1 or 2.
Case 2:
In this case there is a prime $p>n$ with $v_p(m)=s\geq 1$, and division by $m^k$ gives the equation
$$0~=~ \sum \limits_{\ell =0 }^{k-1} {k \choose \ell} \left(\frac{1}{n}\right)^{k-\ell} \,\frac{B_\ell}{m^\ell}.$$
By now setting
$$A_\ell ~=~{k \choose \ell} \left(\frac{1}{n}\right)^{k-\ell} \,\frac{B_\ell}{m^\ell}, $$
one obtains analogous to case 1 that with $\ell<L=\lfloor (k-1)/2\rfloor$
$$\left|A_{2\,L} \right|_p ~\geq~ p^{2\,s\,L},
~~~\left|A_{2\ell} \right|_p \leq p^{2\,s\,\ell+1}< \left|A_{2\ell} \right|_p,$$
such that
$$~\left| \sum \limits_{\ell =0 }^{k-1} A_\ell \right|_p \geq~ p^{2\,s\,L} > 0.$$
This excludes also solutions $x_0=-m/n$ for $n\leq2,~m>n$.
The last remaining case of $x_0=1/2$ is left as exercise for the reader.
