Like my recent question, this one is related to the conjecture stating that, for a prime $p$, the (exact) period of $n\mapsto B_n\bmod p$ is equal to $(p^p-1)/(p-1)$. That question covers a test whether a positive integer $m$ is a period: it suffices to check $B_{m+n}\equiv B_n\pmod{p}$ for $0\leqslant n<p$.
I've found, as it seems, a simpler test (let's denote by $\mathbb{F}_p$ the field $\mathbb{Z}/p\mathbb{Z}$):
An $m>0$ is a period of $n\mapsto B_n\bmod p\ $ if and only if $\ x^p-x-1$ divides $x^m-1$ in $\mathbb{F}_p[x]$.
Compared to this, where I suggest to compute $T=S^m\bmod(S^p-S-1,p)$, and apply it to the initial $p$ values of $B_n\bmod p$ (which require a dedicated computation) with $S$ sending $B_n$ to $B_{n+1}$ (basically, this is a convolution computation), the above suggests just to check $T=1$.
My proof considers the formal power series $\sum_{n=0}^{\infty}B_n x^n$ over $\mathbb{F}_p$, i.e. as an element of $\mathbb{F}_p[[x]]$: \begin{align} \sum_{n=0}^{\infty}B_n x^n&=\sum_{n=0}^{\infty}\prod_{k=1}^{n}\frac{x}{1-kx}=\sum_{q=0}^{\infty}\sum_{r=0}^{p-1}\prod_{k=1}^{pq+r}\frac{x}{1-kx}\\&=\left[\sum_{q=0}^{\infty}\left(\prod_{k=1}^{p}\frac{x}{1-kx}\right)^q\right]\left[\sum_{r=0}^{p-1}\prod_{k=1}^{r}\frac{x}{1-kx}\right]\\&=\left[\sum_{q=0}^{\infty}\left(\frac{x^p}{1-x^{p-1}}\right)^q\right]\left[\sum_{r=0}^{p-1}\prod_{k=1}^{r}\frac{x}{1-kx}\right]\\&=\frac{1-x^{p-1}}{1-x^{p-1}-x^p}\sum_{r=0}^{p-1}\prod_{k=1}^{r}\frac{x}{1-kx}=\frac{N_p(x)}{D_p(x)}, \end{align} where $N_p(x)=\sum\limits_{r=0}^{p-1}x^r\prod\limits_{k=r+1}^{p-1}(1-kx)$ is a polynomial of degree $<p$, and $D_p(x)=1-x^{p-1}-x^p$ is irreducible over $\mathbb{F}_p$ (for this, it suffices to check $D_p(x)\mid x^{p^p}-x$; let's omit the details). On the other hand, if $m$ is a period, then $\sum_{n=0}^{\infty}B_n x^n=(\sum_{n=0}^{m-1}B_n x^n)/(1-x^m)$, and thus $D_p(x)$ divides $(1-x^m)N_p(x)$, hence it divides $1-x^m$. Reflecting, we're done with $\implies$ (the $\impliedby$ is easier).
[We could guess $D_p$ from Touchard's congruence, and in fact don't need $N_p$ explicitly.]
Now the question. Is the above correct? If it is (as I hope), does it bring us closer to resolution?
Update. Found in the second of the articles given in the comment by Gerry Myerson. (At least it is stated there; I didn't track its history back yet.)