Pretty $p^2$-congruences involving Stirling numbers of the both kinds Let  $p$ an odd prime number and ${n\brack {k}}$ (resp. ${n\brace k}$) be the Stirling numbers of first (resp. second) kind, such that:
$$ \sum_{k\ge0} {{n}\brack {k}}x^k = \prod_{j=0}^{n-1}(x+j)$$
$$ \sum_{k\ge0} {{n}\brace {k}}\prod_{j=0}^{k-1}(x-j) = x^n$$
Let $m$ an integer such that $1\le m\le p$, we have 

\begin{align*} {{p+m}\brack {p}}+{{p}\brace {p-m}}&\equiv 0 \pmod
 {p^2}\\ {{p+m}\brace {p}}+{{p}\brack {p-m}}&\equiv 0 \pmod {p^2}
 \end{align*}

This is probably not new. Where can one find a reference? 
EDIT From @i707107 answer, for odd $m$ in the range $3\le m \le p-2 $, we also have the nice 

\begin{align*}{{p+m}\brace {p}}&\equiv {{p}\brack {p-m}} \pmod {p^3}
 \end{align*}

 A: The first congruence
According to the linked paper in comments, we have for $1\leq m<p$, by (1.7)
$$
{{p+m}\brack {p}}\equiv \frac pm \binom{p+m}m (-1)^m B_m^{(m)} \ \textrm{ mod }p^2.$$
Also, by (1.8),
$$
{{p}\brace {p-m}}\equiv -\frac pm \binom{p-1}m B_m^{(m)} \ \textrm{ mod }p^2. 
$$
So, for your first conguence, we need to check if
$$
-\binom{p-1}m+(-1)^m\binom{p+m}m \equiv 0 \ \textrm{ mod }p. 
$$
Writing down the binomial coefficients, we see that their denominators are $m!$ and numerator is zero mod $p$. Thus, your first congruence is true for $1\leq m<p$. 
For $m=p$, the Stirling's second kind gives $0$. The first kind is by (1.4), 
$$
{{2p}\brack {p}}\equiv -\frac{4p^3}{2(p-1)}\binom{2p-1}pB_{p-1} \ \textrm{ mod }p^3.
$$
Let $\nu_p(n)$ be the $p$-adic valuation of $n$. Then we have 
$$
\nu_p(\binom{2p-1}p)=0, \ \ \nu_p(B_{p-1})=-1.
$$
The last one is by von Staudt-Clausen. This gives $p^2|{{2p}\brack {p}}$, the result for $m=p$ follows. 
The second congruence
For even number $m$ in $1\le m <p$, by (1.3) and (1.5),
$${{p}\brack {p-m}}\equiv -\frac pm \binom{p-1}m B_m \ \textrm{ mod } p^2.$$
$$
{{p+m}\brace {p}}\equiv \frac pm \binom{p+m}mB_m \ \textrm{ mod }p^2. 
$$
Thus, the second congruence follows by the same way the first congruence is treated.
For odd number $m$ in $1\le m \le p$, by (1.4) and (1.6),
$${{p}\brack {p-m}}\equiv -\frac{p^2m}{2(m-1)} \binom{p-1}m B_{m-1} \ \textrm{ mod } p^3.$$
$$
{{p+m}\brace {p}}\equiv \frac{p^2m}{2(m-1)} \binom{p+m}m B_{m-1} \ \textrm{ mod } p^3
$$
Also, by von Staudt-Clausen, the Bernoulli number does not have $p$ as a factor of denominator if $m<p$. Thus, we have the result.
For $m=p$, by von Staudt-Clausen, $\nu_p(B_{p-1})=-1$. So, $p^2$ divides both Stirling numbers. The result hence follows. 
