You are indeed missing some pieces. You can immediately see that there is no way your formula works in general since the full expansion reads
$$
\begin{aligned}
e^{ix\cdot P} K_\mu e^{-ix\cdot P} &= \sum_{n,m=0}^\infty \frac{i^{n-m}}{n!m!} (x\cdot P)^n\,K_\mu\,(x\cdot P)^m\,.
\end{aligned}\tag{1}\label{ini}
$$
The problem is that this is a mess because we only know commutators and not products. We need something better. Let me define a useful notation
$$
\mathrm{ad}_X(Y) := [X,Y]\,.
$$
Then I claim that
$$
e^{X}\,Y\,e^{-X} = e^{\mathrm{ad}_X}\,Y\,.\tag{2}
\label{1}
$$
I will prove it later. Now, using this result for $Y= K_\mu$ and $X = ix\cdot P$ we have
$$
e^{ix\cdot P} K_\mu e^{-ix\cdot P} = K_\mu + i x^\nu\,[ P_\nu, K_\mu] - \frac{1}{2!} x^\nu x^\rho [P_\rho,[P_\nu,K_\mu]]\,.
$$
The series would normally go on forever, but this time it terminates because $[P,K]\sim L +D$ and $[P,L]\sim P$, $[P,D]\sim P$. Therefore, going to the next nested commutators would just give $[P_\mu,P_\nu]$, which is zero. Now you can work out the second order nested commutator yourself and see that it matches.
Now let's prove \eqref{1}. For bookkeeping purposes let us define $X\to tX$ where $t \in \mathbb{R}$ and will be set to $1$ at the end. The strategy is to do a Taylor expansion in $t$ and verify that it matches order by order. The subtleties related to whether the Taylor series represents the full function or not will not be addressed in this answer. For order zero we have the obvious equality
$$
e^{0}Y e^{0}= e^{0}Y\,.
$$
Now take a derivative with respect to $t$
$$
e^{tX} \,(XY - YX)\, e^{-tX} = e^{t \,\mathrm{ad}_X} \mathrm{ad}_X(Y)\,,
$$
which is true for $t=0$ by definition of $\mathrm{ad}$. Now call $\mathrm{ad}_X(Y) \equiv Y'$. The second derivative of \eqref{1} just amounts to
$$
e^{tX} \,(XY' - Y'X)\, e^{-tX} = e^{t \,\mathrm{ad}_X} \mathrm{ad}_X(Y')\,,
$$
and clearly they match again. We could continue forever by defining $Y''$ and so on. So the equation \eqref{1} holds order by order and one just has to set $t=1$. Another way to say it is that both sides of the equality satisfy the same first order ODE in $t$ with the same initial conditions, therefore they are equal:
$$
\phi(t) = e^{tX} \,Y\, e^{-tX},\,\qquad \tilde{\phi}(t) = e^{t \,\mathrm{ad}_X} Y\,.
$$
with
$$
\begin{aligned}
\phi(0) &= \tilde{\phi}(0) = Y\,,\\
\phi'(0) &= \tilde{\phi}'(0) = [X,Y]\,.
\end{aligned}
$$
There is also a direct computation that proves it. It starts from my initial formula \eqref{ini} and then uses the following identity
$$
(\mathrm{ad}_X)^m(Y) = \sum_{k=0}^m \binom{m}{k} X^kY(-X)^{m-k}\,,
$$
which you can also prove explicitly. If you want more details, check out Chapter 2 of $[1]$.
$[1]\;$ Hall, Brian C. (2015), Lie Groups, Lie Algebras, and Representations: An Elementary Introduction, Graduate Texts in Mathematics, 222 (2nd ed.), Springer
