Maximum of $E[|X+Z|^m]$ for $Z$ standard normal and $X$ independent of $Z$, two-valued, with $E[|X|^k]=c$ Let $Z$ be a standard normal distribution. 
I am trying to find a solution to the following problem:
\begin{align}
&\max_{ x_1,x_2 \in \mathbb{R}, t\in[0,1]} (1-t) E[|x_1+Z|^m]+t E[|x_2+Z|^m]\\
&\text{ s.t. }  (1-t) |x_1|^k+t|x_2|^k=c
\end{align}
where $0\le m\le k$.
This problem can also be cast as the following problem:
\begin{align}
&\max_{X}  E[|X+Z|^m] \quad (*)\\
&\text{ s.t. }  X \text{ has two mass points}, E[|X|^k]=c, X \text{ is indpendent of } Z
\end{align}
My conjecture is that the above problem is maximized by a deterministic random variable $X= c^{\frac{1}{k}}$ and
\begin{align}
\max_{X} E[|X+Z|^m]= E[|c^{\frac{1}{k}}+Z|^m].
\end{align}
While we are restrict $Z$ to be standard normal it would be nice to have a proof that works for all symmetric and absolutely continuous distributions. 
I feel like the proof should be using Jensen's inequality but not sure how to use it.   The reason is the following. Suppose we remove $Z$ and seek to optimize 
\begin{align}
&\max_{X}  E[|X|^m]\\
&\text{ s.t. }  X \text{ has two mass points}, E[|X|^k]=c, X \text{ is indpendent of } Z
\end{align}
Since, $m \le k$ by Jensen's inequality
\begin{align}
E[|X|^m] \le (  E[|X|^k]  )^{\frac{m}{k}}.
\end{align}
Note, that Jensen's inequality is equality iff $X$ is a constant.  So, the optimization problem $\max_{X}  E[|X|^m]$ is solve by deterministic random variable. 
Edit 1:  It appears that my conjecture is only true in some cases. See a very nice approach of kimchilover.
Edit 2:  It also appears that in $(*)$ the $\max$ should be replaced with $\sup$. This was also pointed out by kimchilover.
 A: This problem is harder, and more interesting, than I originally thought. The OP's conjecture is sometimes true and sometimes not, depending on the values $k$ and $m.$
For given $m$ let $g(x) = E|Z+x|^m,$ where $Z\sim   N(0,1).$ The symmetry of the distribution of $Z$ implies $g(x)=g(-x)$. If $m\ge 1$ the function $g$ is convex on $\mathbb R.$  But $g$ is concave on $[0,\infty)$ if $m<1$.   As outlined in a comment, let $\Gamma=\{(x^k,g(x)): x\ge 0\} = \{(u, g(u^{1/k})): u\ge 0\}.$  (Note the comment did not restrict $x\ge0,$ but it should have.) Let $H\subset \mathbb R^2$ be the  closed convex hull of $\Gamma.$ The closure of the set of possible $(E|X|^k, E|X+Z|^m)$ values attainable as we vary the probability law of $X$ is exactly the set $H$.  Let $L=\{(x,y)\in\mathbb R^2: x=c\}$. The intersection $H\cap L$ is a closed interval in the plane; the OP's question concerns the upper end point of that interval.
There are two cases to consider.  In the easy case, the function $\gamma: u\mapsto g(u^{1/k})$ is concave.  Then the desired maximum is $\gamma(c^{1/k})$, attained at $(c,g(c))\in\Gamma$ when $P(X=c^{1/k})=1$, as the OP conjectured.
In the harder case $g$ is not concave.  Let $\gamma^*$ be the upper concave envelope of $\gamma$, that is, the pointwise infimum of all concave functions pointwise greater than $\gamma$.  By Dubins's theorem, or by Rockafellar's Corollary 17.1.5 (transposing his convex to our concave, etc.) we know that $(c,\gamma^*(c))$ is a convex combination of at most two points on $\Gamma$, which is what the original problem statement wanted. Note also that this $(c,\gamma^*(c))$ is in the closure of the hull of $\Gamma$ and itself might be only a limit of 2 point mixtures, not a 2 point mixture itself.
And there might be technical conditions needed to apply these theorems.
So what distinguishes the easy case from the hard?  First, if $m$ is an even integer, $g(x)$ works out to an even  polynomial with nonnegative coefficients, of degree $m$.  Thus $\gamma(u)$ is a nonnegative  combination of terms $u^{j/k}$ for $0\le j\le m$, each of which is concave, so long as $k\ge m$.  Another case to consider is $m=1.$  Then it can be checked that $\gamma$ is not concave unless $k\ge 2$.  In general, for $\gamma$ to be concave  the function $u\mapsto g'(u^{1/k})u^{1/k-1}$ must be decreasing in $u\in\mathbb R_+$) which is to say, $g'(x)x^{1-k}$ is decreasing in $x\in \mathbb R_+$.
ADDED 26 July.  In the non concave case the sought-after maximum of $E|X+Z|^m$ need not be attained, but only approximated by 2-point distributions for $X$.  In the $k=m=1$ case the function $\gamma$ is convex and increasing, and the set $H$ is $\{(x,y): x\ge 0, g(x)\le x \le x+g(0)\},$ but the points of form $(x,x+g(0))$ for $x>0$ are not in the convex hull of $\Gamma$ but only in its closure.  They cannot be attained by any distribution on $X$.  Thus the original statement of the problem should have  asked for $\sup E|X+Z|^m$ instead of $\max E|X+Z|^m.$
ADDED & edited (to cope with $k\ne1$) 29 July.  In hindsight I could have picked notations a bit better. Let $\gamma(x) = g(x^{1/k}).$
Let $\Gamma=\{(x,\gamma(x)): x\ge0\}$ be the graph of $\gamma$.  Let $S$ be the convex hull of $\Gamma$, and let $\overline S$ be the closure of $S$.  Let $\gamma^*$ be the upper concave envelope of $\gamma$. The problem of optimizing $E\gamma(X)$ over all $X\ge$ such that $E|X|^k=c$ is solved by $\gamma^*(c)$ in one of these senses:  If $(c,\gamma^*(c))\in \Gamma$ the optimum is attained, as the OP conjectured, by a 1-point distribution for which $P(X=c)=1.$  If instead $(c,\gamma^*(c))\in S \setminus \Gamma,$ the optimum is attained by a 2-point distribution.  But if $(c,\gamma^*(c))\in \overline S \setminus S,$ the optimum is not attained, but only approximated by 2-point distributions.
Added 3 August:  The example $m=k=1$ is instructive.  One finds that $g(x)= E|Z+x| = 2\varphi(x) + x(2\Phi(x)-1)$, a convex function on $[0,\infty)$.  (Here $\varphi$ and $\Phi$ are the standard normal density and c.d.f.)  The upper concave envelope is evidently   $g^*( x) = g(0)+x.$ The convex hull, $S,$ of the graph $\Gamma=\{(x,g(x)): x\ge 0\}$ is given by $S=\Gamma \cup \{(x,y): x>0, g(x)\le y < g(0)+x\}$; it is seen to be the union of $\Gamma$ and all chords connecting pairs of points on $\Gamma$; it is also the set of all pairs $(EX,Eg(x))$ attainable by 1 or 2 point mass distributions for $X$. The point $(1,g(0)+1)$ is not in $S$, and cannot be represented by two point masses.   The sequence of 2 point distributions $\mu_n = (1-1/n)\delta_0 + \delta_n/n$ obeys $\int x\mu_n(dx) = 1$ and $\int g(x) \mu_n(dx) = (1-1/n)g(0)+ 2\phi(n)/n + n(2\Phi(n)-1)/n \to g(0)+1$.  (Claims in the comments to the effect that the set of measures $\mu$ such that $E|x|\le c$ is compact are besides the point: because $\mathbb R_+$ is unbounded, the map $\mu\mapsto \int x \mu(dx)$ is not continuous.  The sequence $\mu_n$  might converge weakly to $\delta_0$ but the sequence $\int g(x) \mu_n(dx)$ goes not converge to $g(0)$.)
