$\sup_{x\in\Bbb R}|F(x+iy)|\leq M_0^{1-y}M_1^y$ on the strip $0$\newcommand{\ep}{\epsilon}$
The following is Problem 4.3 in Stein complex analysis

Suppose $F(z)$ is holomorphic and bounded in the strip $S = \{z:0<Im(z)<1\}$ and continuous on its closure. If $|F(z)|\leq 1$ on the boundary lines, then $|F(z)|\leq 1$ throughout the strip.
Question. For the more general $F$, let $\sup_{x\in\Bbb R}|F(x)| = M_0$ and $\sup_{x\in\Bbb R}|F(x+i)| = M_1$. Then,
$$\sup_{x\in\Bbb R}|F(x+iy)|\leq M_0^{1-y}M_1^y,\quad 0\leq y\leq 1.$$
[Hint: Consider $M_0^{z-1}M_1^{-z}F(z)$.]

Let $G(z) =M_0^{z-1}M_1^{-z}F(z)$ and $G_{\ep} = M_0^{z-1}M_1^{-z}F(z)e^{-\ep z^2}$. Then as $F$ is bounded, $|F(z)|\leq M$ on $S$. Hence, if $z = x+iy$ then
\begin{align*}
|G(z)|& \leq |M_0^{z-1}M_1^{-z}M|\\
& \leq |M_0^{x-i+iy}M_1^{-x-iy}M|\\
& = M_0^{x-1}M_1^{-x}M\\
|G_{\ep}(z)|&\leq M_0^{x-1}M_1^{-x}Me^{-\ep(x^2-y^2)}\\
& = M'e^{-\ep x^2+x(\log M_0 -\log M_1) - \log M_0}\to 0,\quad |x|\to\infty.\\
\end{align*}
So by MMP, $G_{\ep}$ is bounded by its boundary bounds.
\begin{align*}
|G(x)| & = |F(x)M_0^{x-1}M_1^{-x}|\\
|G(x)| & \leq (M_0/M_1)^x\\
|G(x+i)| & = |F(x+i)M_0^{x-1+i}M_1^{-x-i}|\\
&\leq (M_0/M_1)^{x-1}.\\
\end{align*}
I'm stuck here. How can I get the desired bound from this?
 A: Here's a proof done via symmetrization, combined with some ideas from real analysis.
The Lemma above the bolded question in the OP really states that

Suppose $F(z)$ is holomorphic and bounded in the strip $S' = \big\{z:a<Im(z)<b\big\}$, for real $a\lt b$, and continuous on its closure. If $|F(z)|\leq K$ on the boundary lines, then $|F(z)|\leq K$ throughout the strip

Now for non-negative integers $n$
$E_n:= \big\{\frac{m}{2^n}: m=0,1,2,3, \dots, 2^{n}\big\}$, noting that
$E_n\subset E_{n+1}$ and between two consecutive point of $E_n$ there is exactly one point of $E_{n+1}$
For reasons of continuity it's enough to prove your desired bound that for $x+iy \in \overline S = \big\{z:0\leq Im(z)\leq 1\big\}$
$\big \vert F(x+iy)\big \vert\leq M_0^{1-y}M_1^y,\quad 0\leq y\leq 1$
for each $y\in E_n$ for arbitrary $n$ (and the general result holds by taking limits)
as an aside: we may assume $M_0,M_1 \gt 0$, as e.g. $M_0=0$ readily implies that $f$ is identically zero (why?).

Proceed via induction on $k$.
Base case
The statement holds trivially for all elements in $E_0$, for arbitrary $x\in \mathbb R$.
Inductive case:
The result holds for all elements in $E_{k}$ and we need to show it holds for all elements in $E_{k+1}$.
For avoidance of doubt: the inductive  hypothesis is that the desired bound holds for arbitrary $x\in\mathbb R$, for all $y\in E_k$ for any map $:\overline S\longrightarrow \mathbb C$, that is analytic on $S$, and continuous on the boundary with upper bounds $M_0$ and $M_1$ on the respective boundary lines.
(i.) specialize to $x=0$
Select $y:=\frac{m}{2^{k+1}}\in E_{k+1}$ and if $y\in E_{k}$ there is nothing to do, so assume $y\not\in E_{k}$ (i.e. that $m$ is odd).  Set $a:= y-\frac{1}{2^{k+1}}$ and $b:= y+\frac{1}{2^{k+1}}$ and we have an analytic function $f$ given by $f(z)=F(z)$ in the strip $S' = \big\{z:a<Im(z)<b\big\}$, continuous on the boundary lines (i.e. $f$ is the restriction of $F$ to $\overline S'$) with respective upper bounds $M_0'= \Big(M_0^\frac{m-1}{2^{k+1}}M_1^{1-\frac{m-1}{2^{k+1}}} \Big)$ and $M_1'=\Big(M_0^\frac{m+1}{2^{k+1}}M_1^{1-\frac{m+1}{2^{k+1}}} \Big)$, by induction hypothesis
$g\big(z\big):=f\big(z\big)\cdot f\big((a+b)i-z\big)=f\big(0+iy\big)\cdot f\big(0+i(a+b-y)\big)$
where $g$ has a symmetrized upper bound of $K:=M_0'\cdot M_1'$ on each of the boundary lines
and at the midpoint:
$\Big \vert f\big(0+i\frac{a+b}{2}\big)\Big \vert^2 = \Big\vert g\big(0+i\frac{a+b}{2}\big)\Big \vert\leq M_0'\cdot M_1' = \Big(M_0^\frac{m-1}{2^{k+1}}M_1^{1-\frac{m-1}{2^{k+1}}} \Big)\cdot\Big(M_0^\frac{m+1}{2^{k+1}}M_1^{1-\frac{m+1}{2^{k+1}}} \Big)$
where the upper bound comes via application of the Lemma.  Taking square roots:
$\implies f\Big\vert\big(0 +i y\big)\Big\vert = f\Big\vert\big(0 +i \frac{m}{2^{k+1}}\big)\Big\vert=\Big \vert f\big(0+i\frac{a+b}{2}\big)\Big \vert \leq M_0^\frac{m}{2^{k+1}}\cdot M_1^{1-\frac{m}{2^{k+1}}}$
which completes part (i) of the inductive step.  That is, we've proven the inductive step in the special case of $x=0$ for any map $:\overline S\longrightarrow \mathbb C$, that is analytic on $S$, and continuous on the boundary with upper bounds $M_0$ and $M_1$ on the respective boundary lines.
(ii.) now in general select and fix arbitrary $x'\in \mathbb R$ and consider translation
$T_{x'}:\overline S\longrightarrow \overline S$ given by
$T_{x'}\big(x+iy\big) = (x+ x')+ iy$
$h:\overline S\longrightarrow \mathbb C$ given by $h:=f\circ T_{x'}$
$h$ is analytic in $S$, continuous on the two boundary lines with respective upper bounds of $M_0$ and $M_1$.  Then for $x'+iy \in S$
$f(x'+iy)=h(0+iy)$, and (i.) applied to $h$ shows that for $y\in E_{k+1}$
$\big \vert f(x'+iy)\big \vert=\big \vert h(0+iy)\big \vert\leq M_0^\frac{m}{2^{k+1}}\cdot M_1^{1-\frac{m}{2^{k+1}}}$
which completes the induction and the proof
