I have 2 comments:
Empirically I have a guess for the positive real roots: Let a scaling-factor $c=\frac1{\ln 2}-1 \approx 0.442$ then $$ \rho_k \approx (k+3) \cdot c $$ or $$ k \lt {\rho_k \over c }-3 \lt k+1 \tag {for $k \ge$ 5}$$ I checked this up to $k=5000$ so far using the bernoulli-polynomials and internal float precision of 400 digits in Pari/GP (see table below).
looking at the equation modulo $k$ it seems, the lhs of the equation is always equivalent to zero, so solutions can only exist, if also the rhs is zero modulo $k$.
Let $w$ be the squarefree kernel of primefactors of $k$. Then the rhs is equivalent zero only if $x$ is a multiple of $w$ and thus only such $x$ can solve the equality.
I hoped to combine the two empirical results somehow, but don't see any further useful property - so also I do not know, whether it is at all worth the effort to actually prove observation 1).
Here is a list of the roots, scaled by the scaling factor and the integer value $3$ and the fractional value separated: $$\small \begin{array} {r|l} k & {\rho_k \over c} \\ \hline 4 & 4 + 4.16684831867 \\ 5 & 5 + 3.65764105650 \\ 6 & 6 + 3.48606997990 \\ 7 & 7 + 3.4241172248 \\ 8 & 8 + 3.4015117133 \\ 9 & 9 + 3.3933253612 \\ 10 & 10+3.3903889002 \\ 11 & 11+3.3893375033 \\ 12 & 12+3.3889540764 \\ 13 & 13+3.3888054311 \\ 14 & 14+3.3887395158 \\ 15 & 15+3.3887034724 \\ \ldots & \ldots \\ 497 & 497+3.388349245 \\ 498 & 498+3.388349220 \\ 499 & 499+3.388349196 \\ 500 & 500+3.388349172 \end{array} $$