I have 2 comments:
1) 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).
2) 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 $k$ - here I assume only odd $k$ so far.
Let $w$ be the squarefree kernel of primefactors (all are odd) 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).
<hr>
Here is a list of the roots, scaled by the scaling factor and the integer value $3$ and the fractional value separated:
$$\small \begin{array} {r|l}
k & {\rho_k \over c} & {\rho_k \over c}=k+{1.5 \over c}+ \epsilon \\
\hline
4 & 4 + 4.16684831867 & 4 + 1.5/c + 0.344643086811 \\
5 & 5 + 3.65764105650 & 5 + 1.5/c + 0.119219557064 \\
6 & 6 + 3.48606997990 & 6 + 1.5/c + 0.0432658922944 \\
7 & 7 + 3.4241172248 & 7 + 1.5/c + 0.0158397148376 \\
8 & 8 + 3.4015117133 & 8 + 1.5/c + 0.00583236702541 \\
9 & 9 + 3.3933253612 & 9 + 1.5/c + 0.00220830950877 \\
10 & 10+3.3903889002 & 10 + 1.5/c + 0.000908352824271 \\
11 & 11+3.3893375033 & 11 + 1.5/c + 0.000442904593529 \\
12 & 12+3.3889540764 & 12 + 1.5/c + 0.000273163432055 \\
13 & 13+3.3888054311 & 13 + 1.5/c + 0.000207358901590 \\
14 & 14+3.3887395158 & 14 + 1.5/c + 0.000178178523631 \\
15 & 15+3.3887034724 & 15 + 1.5/c + 0.000162222283615 \\
\ldots & \ldots \\
497 & 497+3.388349245 & 497 + 1.5/c + 0.00000540736229828 \\
498 & 498+3.388349220 & 498 + 1.5/c + 0.00000539654486254\\
499 & 499+3.388349196 & 499 + 1.5/c + 0.00000538577062079\\
500 & 500+3.388349172 & 500 + 1.5/c + 0.00000537503931484\\
\end{array}
$$
<hr> [update]
Here is another list, where the exponent $k$ is in exponential progression and the remaining ${\epsilon \over c}$ (from the above table) is displayed by its base-$2$-logarithm:
$$\small{ \begin{array} {rr|l}
j & k=2^j & \log_2 (\epsilon /c) \\
\hline\\
2 & 4 & -0.361210135710 \\
3 & 8 & -6.24608789212 \\
4 & 16 & -11.5139506158 \\
5 & 32 & -12.4418890639 \\
6 & 64 & -13.4004277751 \\
7 & 128 & -14.3795231280 \\
8 & 256 & -15.3690269163 \\
9 & 512 & -16.3637677898 \\
10 & 1024 & -17.3611354651 \\
11 & 2048 & -18.3598186117 \\
12 & 4096 & -19.3591600122 \\
13 & 8192 & -20.3588306692 \\
14 & 16384 & -21.3586659868 \\
15& 32768 & -22.3585836430\\
\end{array} }$$
*(The range of $k$ was not continuously computed, only at the explicite values)*