Do there exist an infinite number of complex solutions of $3^z+4^z=5^z$? Are the followings true?
1 : There exist an infinite number of complex solutions of the equation 
$$3^z+4^z=5^z \tag{$\star$}.$$
2 : There exist an infinite number of complex solutions $z$ of $(\star)$ such that $2-\varepsilon\lt Re(z)\lt 2+\varepsilon$ for any $\varepsilon\gt0.$
Motivation : 
It's easy to prove that $x=2$ is the only one real solution of $(\star)$. Then, I got interested in finding complex solutions of $(\star)$. Can anyone help?
 A: Summary
For the equation $3^z + 4^z = 5^z$,


*

*$z = 2$ is the only root when $\Re z \ge 2$.

*There are infinitely many complex roots within the strip: $\;\;2 - \varepsilon < \Re z < 2$.


Details
Let $u, v$ be two positive numbers such that $u^2 + v^2 = 1$.
Let $\mu = \frac{|\log u|}{2\pi}, \nu = \frac{|\log v|}{2\pi}$ and
consider following Dirichlet polynomial equation:
$$\varphi(z) = u^z + v^z = 1\tag{*1}$$


*

*By construction, $z = 2$ is a root of $(*1)$. 

*For $\Re z > 2$, we have:
$$|u^z+v^z| \le |u^z| + |v^z| = u^{\Re z} + v^{\Re z} < u^2 + v^2 = 1.$$
This means $(*1)$ doesn't have any root with $\Re z > 2$.

*For $\Re z = 2$, let $z = 2 + it$, we have:
$$\begin{align}
u^z+v^z = 1 
\iff & u^2 e^{-2\pi\mu t i} + v^2 e^{-2\pi\nu t i} = 1\\
\iff & e^{-2\pi\mu t i} = e^{-2\pi\nu t i} = 1\\
\iff & \mu t, \nu t \in \mathbb{Z}
\end{align}$$
For $t \ne 0$, it is clear the last condition $\mu t, \nu t \in \mathbb{Z}$ is possible
when and only when $\frac{\mu}{\nu}$ is a rational number. This means:


*

*if $\frac{\mu}{\nu} \in \mathbb{Q}$, $(*1)$ has infinitely many roots on the line $\Re z = 2$.

*if $\frac{\mu}{\nu }\notin \mathbb{Q}$, $z = 2$ is the only root of $(*1)$ with $\Re z \ge 2$.



Let us now concentrate on the more interesting case where $\frac{\mu}{\nu} \notin \mathbb{Q}$. We know there are infinitely many pairs of positive integers $p,q$ such
that
$$\left|\frac{\mu}{\nu} - \frac{p}{q}\right| < \frac{1}{q^2}$$
For any such pair of integers $p, q$, let 
$t_{\mu} = \frac{p}{\mu}$, $t_{\nu} = \frac{q}{\nu}$, $t_{<} = \min(t_\mu,t_\nu)$, and $t_{>} = \max(t_\mu,t_\nu)$.
When $t$ varies between $t_{<}$ and $t_{>}$, we have
$$\begin{align}
& |\mu t - p | \le |\mu t_{\nu} - p| = | \frac{\mu}{\nu} q - p | < \frac{1}{q}\\
& |\nu t - q | \le |\nu t_{\mu} - q| = | \frac{\nu}{\mu} p - q | < \frac{\nu}{\mu q}\tag{*2}
\end{align}$$
Let $K = \max(\frac{\nu}{\mu},1)$, one consequence of $(*2)$ is when $q > 4K$, the angles of following two circular arcs on unit circle
$$
C_\mu = \Big\{ \omega = e^{-2\pi\mu t i} : t \in [t_<, t_>]\Big\}
\quad\text{ and }\quad
C_\nu = \Big\{ \omega = e^{-2\pi\nu t i} : t \in [t_<, t_>]\Big\}
$$
are both smaller than $\frac{\pi}{2}$. 
Since $C_{\mu}$ touches $1$ at and only at $t = t_\mu$ and
$C_{\nu}$ touches $1$ at and only at $t = t_\nu$,
One of $C_{\mu}$ or $C_{\nu}$ ( $C_\mu$ if $t_\mu > t_\nu$, $C_\nu$ otherwise ) 
lies completely in the quadrant  $\Re \omega > 0, \Im \omega \ge 0$ 
while the other one lies completely in the quadrant $\Re \omega > 0, \Im \omega \le 0$.
Since $C_\mu$ and $C_\nu$ touches $1$ at different $t$, we can deduce
$0 < \Re \varphi(2+it) < 1$ for $t \in (t_<, t_>)$. Furthermore, it is not hard 
to see $\Im \varphi(2 + it_<) > 0$ and $\Im \varphi(2 + it_>) < 0$.
Let $z_< = 2 + i t_<$, $z_> = 2 + i t_>$ and $m = \min(\mu,\nu)$.
In addition to $q > 4K$, let us assume we have chosen a $q$ so large such that
$$e^{2\pi m\varepsilon}\cos(\frac{2\pi K}{q}) > 1.$$
Let $\mathscr{R}$ be the 
rectangular contour joining $z_<$, $z_>$, $z_>\!-\varepsilon$, $z_<\!-\varepsilon$ and then back to $z$. We will look at what happens to $\omega = \varphi(z) - 1$ when $z$ walks around
$\mathscr{R}$ once.


*

*Along the line segment $[z_<,z_>]$, above discussion implies 
$\omega$ moves from the quadrant $\Re \omega < 0, \Im \omega > 0$ to 
the quadrant $\Re \omega < 0, \Im \omega < 0$.

*Along the line segment $[z_>, z_>\!-\varepsilon]$, it is easy to see
$\Im\omega$ doesn't change sign and remains $< 0$ all the time. 

*Along the line segment $[z_>\!-\varepsilon, z_<\!-\varepsilon]$, we have
$$\Re\omega \ge 
u^2 e^{2\pi\mu\varepsilon}\cos(\frac{2\pi}{q}) + 
v^2 e^{2\pi\nu\varepsilon}\cos(\frac{2\pi\nu}{\mu q}) - 1
\ge e^{2\pi m\varepsilon}\cos(\frac{2\pi K}{q}) - 1
> 0
$$

*Along the line segment $[z_<\!-\varepsilon, z_<]$, $\Im\omega$ 
doesn't change sign again and remains $> 0$ all the time.
Combine these 4 observations, we can conclude when $z$ walks around $\mathscr{R}$ once,
$\omega$ walks around the origin counter-clockwisely once. From this, we can conclude
$\varphi(z) = 1$ has a root inside $\mathscr{R}$. Since there are infinitely many $q$ one
can chose from, we conclude $\varphi(z) = 1$ has infinitely many roots in the strip
$2 - \varepsilon < \Re z < 2$.
To answer the original question, let $u = (3/5)^2$ and $v = (4/5)^2$. 
It is known that $\log 2$, $\log 3$ and $\log 5$ are linearly independent over $\mathbb{Q}$.
This means $\frac{\mu}{\nu} = \frac{\log u}{\log v} = \frac{\log 5 -  \log 3}{\log 5 - 2\log 2} \notin \mathbb{Q}$.
By above discussions, the conclusions in the summary follow immediately.
A: It is a consequence of Hadamard's factorization theorem that any entire function of order $1$ with finitely many zeros is of the form $e^{az}p(z)$, where $a \in \mathbb{C}$ and $p(z)$ is a polynomial.  The function $f(z) = 3^z + 4^z - 5^z$ is entire and of order $1$ but is not of this form, so it must have infinitely many zeros.
A: Not an answer, but you may be interested in looking at the phase plot of $f(z) = 3^z+4^z-5^z$.  You could do so here: http://www.math.osu.edu/~fowler.291/phase/.  You can zoom out using a scroll wheel, or 2 fingers on a touchpad.
It certainly looks like there are infinitely many zeros!  I will edit this answer if I think of a way to show that.
