If HP=PH+P for H,P n×n complex matrices, must H be diagonalizable? 
If $F$ is a field of characteristic zero, $H,P$ are $n\times n$ matrices over $F$, $0 \neq \alpha \in F$, and $HP=PH+\alpha P$, then must the minimal polynomial of $H$ be square-free and must $P$ be nilpotent?

Progress:
It suffices to consider $F$ algebraically closed. I first tried to find counterexamples, but failed. I then showed $P$ is nilpotent:
For any polynomial $f \in F[x]$, one has $f(H)\cdot P = P \cdot f(H+\alpha)$. This means $P$ moves $H$-invariant subspaces to (usually other) $H$-invariant subspaces.
Let $m_H(x)$ be the minimal polynomial of $H$. Then $f(H)$ has a nonzero element of its null-space if and only if $f(x)$ and $m_H(x)$ have a (non-unit) common factor.
Let $\vec{v} \in F^n$ be some nonzero vector. Let $k=k(\vec{v})$ be the least positive integer so that $P^k \vec{v} = \vec{0}$ (or $k=\infty$ if no such $k$ exists). Set $g_i(x) = m_H(x-i\alpha)$. Since $P^i f(H) = f(H-i\alpha) P^i$ for all non-negative integers $i$, we get that $P^i \vec{v} \in \ker( g_i(H) )$. There can be only finitely many $i$ so that $\gcd(m_H(x),m_H(x-i\alpha))\neq 1$ since the set $\{i\alpha : i =0,1,\ldots\}$ is infinite when $F$ has characteristic zero. Hence there must be at least one $i$ with $\ker( g_i(H))=0$, and so with $P^i \vec{v} = \vec{0}$. Hence $k < \infty$. Clearly $P^i \vec{v} = \vec{0}$ for all $i \geq k$.
Now just take the maximum of the $k(\vec{v})$ for $\vec{v}$ in a basis to see $P^k  = 0$ is the zero matrix.
Progress under increased hypotheses:
If $F[H,P]$ acts irreducibly on $F^n$, then letting $f(x)$ be the square-free part of $m_H(x)$, we find that $\ker(f(H))$ is both $H$- and $P$-invariant, so $f=m_H$.
If there is another matrix $M$ with $[H,M]=-\alpha M$ and $[P,M]=H$, then some results in Lie algebras show $H$ is diagonalizable (without requiring the action to be irreducible, right?).
Bonus:
Only available if the initial question is answered (show $m_H(x)$ is square-free or give a counterexample).

Is anything salvageable when $F$ has characteristic $p$?

Even if $F$ is algebraically closed, the set $\{i\alpha:i=0,1,\ldots\}$ has order $p$ so is finite. Is $P$ still nilpotent? Under the increased hypotheses, things still work, I think, but I am suspicious that those hypotheses magically fix the characteristic dependence.
 A: For $n \geq 4$ this is false. For $n=2,3$ I'm still not sure (although leaning toward it being true).
$$H = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$
$$P = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$
Then $[H,P]=2P$ ($P$ is nilpotent) but $H$ is not diagonalizable.
This really felt like it should be true. $H$ feels like a basis element for a Cartan subalgebra of $\mathfrak{sl}_2$ and $P$ a root vector. If we could find a proper "transpose" of sorts for $P$ we could get a copy of $\mathfrak{sl}_2$ and $H$ would necessarily be diagonalizable. 
I don't have any sort of proof, but I think $\mathfrak{gl}_2$ and $\mathfrak{gl}_3$ are too "small" to accommodate a counter-example. I believe your statement is true there...but alas I can't seem to see how to prove it.
A: I realized my classification of the $F[H,P]$ representations (which assumed $H$ diagonalizable) actually worked for any Jordan block, not just a $1\times 1$.
Classification with $H$ diagonalizable:
The idea is that $P$ bumps the eigenspaces of $H$. This works even if the eigenspaces are generalized. The classic example is $\mathfrak{b} \leq \mathfrak{sl}_2$ with $H=\begin{bmatrix}1 & 0 \\ 0& -1 \end{bmatrix}$ and $P=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ and $\alpha=2$. The classification is $H$ is a direct sum of diagonal matrices $H_n(\lambda)$ with diagonal entries $\lambda, \lambda - \alpha, \lambda-2\alpha, \ldots, \lambda - (n-1) \alpha$ and $P$ is a direct sum of nilpotent matrices $P_n(\lambda)$ that are Jordan blocks of size $n$ with eigenvalue $0$.
Example with $H$ not diagonalizable
However, if $H$ is not required to be diagonalizable, then $\lambda$ doesn't have to be a number. It can be any matrix.
Here is another example with $H$ not diagonalizable:
$$
H=H_2\left(\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix}\right) 
= \left[\begin{array}{rr|rr}
1 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 \\ \hline 
0 & 0 & -1 & 1 \\
0 & 0 & 0 & -1
\end{array}\right],\quad
P=P_2\left(\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix}\right)
= \left[ \begin{array}{rr|rr}
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\ \hline
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{array}\right]$$
Classification:
More generally $H_n(A)$ and $P_n(A)$ for any non-diagonalizable matrix $A$ works. Here $H_n$ is a block diagonal matrix with blocks $A, A-\alpha I_n, A-2\alpha I_n, \ldots, A-(n-1)\alpha I_n$ and $P_n(A)$ is a nilpotent block matrix all of whose blocks are the same size as $A$, most of which are zero, except for those on the first super diagonal which are $I_n$. To avoid duplicates, one should probably require $A$ to be a Jordan block.
Trivial counterexample:
Sillily enough, one can even take $n=1$! Let $H$ be any non-diagonalizable matrix, and let $P=0$. Oops! This is why the $[P,M]=H$ relation of $\mathfrak{sl}_2$ is so important.
