The following shows how to transform the quartic
\begin{equation}
y^2=ax^4+bx^3+cx^2+dx+e ,
\end{equation}
with $a,b,c,d,e \in \mathbb{Q}$, into an equivalent elliptic curve.
Case 1:
We first consider the case when $a=1$, which is very common. If $a=\alpha^2$ for some rational $\alpha$, we
substitute $y=Y/\alpha$ and $x=X/\alpha$, giving
\begin{equation*}
Y^2=X^4+\frac{b}{\alpha}X^3+cX^2+\alpha d X+\alpha^2 e
\end{equation*} Thus, suppose
\begin{equation}
y^2=x^4+bx^3+cx^2+dx+e
\end{equation}
We describe the method given by Mordell on page $77$ of his book Diophantine Equations, with some minor modifications.
We first get rid of the cubic term by making the standard substitution $x=z-b/4$ giving
\begin{equation*}
y^2=z^4+fz^2+gz+h
\end{equation*}
where
\begin{equation*}
f=\frac{8c-3b^2}{8} \hspace{1cm} g=\frac{b^3-4bc+8d}{8} \hspace{1cm}
h=\frac{-(3b^4-16b^2c+64bd-256e)}{256}
\end{equation*}
We now get rid off the quartic term by defining $y=z^2+u+k$, where $u$ is a new variable and $k$ is a constant to be
determined. This gives the quadratic in $z$
\begin{equation*}
(f-2(k+u))z^2+gz+h-k^2-u(2k+u)=0
\end{equation*}
For $x$ to be rational, then $z$ should be rational, so the discriminant of this quadratic would have to be a rational square.
The discriminant is a cubic in $u$. We do not get a term in $u^2$ if we make $k=f/6$, giving
\begin{equation*}
D^2=-8u^3+2\frac{f^2+12h}{3}u+\frac{2f^3-72fh+27g^2}{27}
\end{equation*}
and, if we substitute the formulae for $f,g,h$, and clear denominators we have
\begin{equation}
G^2=H^3+27(3bd-c^2-12e)H+27(27b^2e-9bcd+2c^3-72ce+27d^2)
\end{equation}
with
\begin{equation}
x=\frac{2G-3bH+9(bc-6d)}{12H-9(3b^2-8c)}
\end{equation}
and
\begin{equation}
y= \pm \frac{18x^2+9bx+3c-H}{18}
\end{equation}
Case 2:
If $a \ne \alpha^2$, we need a rational point $(p,q)$ lying on the quartic.
Let $z=1/(x-p)$, so that $x=p+1/z$ giving
\begin{equation*}
y^2z^4=(ap^4+bp^3+cp^2+dp+e)z^4+(4ap^3+3bp^2+2cp+d)z^3+
\end{equation*}
\begin{equation*}
(6ap^2+3bp+c)z^2+(4ap+b)z+a
\end{equation*}
where the coefficient of $z^4$ is, in fact, $q^2$. Define, $y=w/z^2$, and then
$z=u/q^2, \, w=v/q^3$
giving
\begin{equation}
v^2=u^4+(4ap^3+3bp^2+2cp+d)u^3+q^2(6ap^2+3bp+c)u^2+
\end{equation}
\begin{equation*}
q^4(4ap+b)u+aq^6 \equiv u^4+fu^3+gu^2+hu+k
\end{equation*}
We now, essentially, complete the square. We can write
\begin{equation*}
y^2=u^4+fu^3+gu^2+hu+k=(u^2+mu+n)^2+(su+t)
\end{equation*}
if we set
\begin{equation*}
m=\frac{f}{2} \hspace{1cm} n=\frac{4g-f^2}{8} \hspace{1cm} s=\frac{f^3-4fg+8h}{8}
\end{equation*}
and
\begin{equation*}
t=\frac{-(f^4-8f^2g+16(g^2-4k))}{64}
\end{equation*}
This gives
\begin{equation*}
(y+u^2+mu+n)(y-u^2-mu-n)=su+t
\end{equation*}
and if we define $y+u^2+mu+n=Z$ we have
\begin{equation}
2(u^2+mu+n)=Z-\frac{su+t}{Z}
\end{equation}
Multiply both sides by $Z^2$, giving
\begin{equation*}
2u^2Z^2+2muZ^2+suZ=Z^3-2nZ^2-tZ
\end{equation*}
which, on defining $W=uZ$, gives
\begin{equation*}
2W^2+2mWZ+sW=Z^3-2nZ^2-tZ
\end{equation*}
Define, $Z=X/2$ and $W=Y/4$ giving
\begin{equation}
Y^2+2mXY+2sY=X^3-4nX^2-4tX
\end{equation}
which is an elliptic curve. If we define $Y=G-s-mX$, we transform to the form
\begin{equation}
G^2=X^3+(m^2-4n)X^2+(2ms-4t)X+s^2
\end{equation}
All of the above are easy to program in any symbolic package. Hopefully, I have all the formulae correct.
It's an interesting question.
For the particular equation
$$
3^x+5^z=4^y
\qquad\qquad\;\;\;\,
$$
your attempt to prove that there are no positive integer solutions had flaws, as was noted in the comments.
However for that equation, there are, in fact, no positive integer solutions.
This can be shown as follows . . .
Let $\mathbb{N}$ denote the set of positive integers and let $R,S,T$ be defined as
\begin{align*}
R&=\{3^x\;\text{mod}\;20{\,\mid\,}x\in\mathbb{N}\}
\\[4pt]
S&=\{5^z\;\text{mod}\;20{\,\mid\,}z\in\mathbb{N}\}
\\[4pt]
T&=\{4^y\;\text{mod}\;20{\,\mid\,}y\in\mathbb{N}\}
\\[4pt]
\end{align*}
Then we get
\begin{align*}
R&=\{1,3,7,9\}
\qquad\qquad\;\,
\\[4pt]
S&=\{5\}
\\[4pt]
T&=\{4,16\}
\\[4pt]
\end{align*}
and it's then easily verified that there do not exist $r,s,t$ with $r\in R,\;s\in S,\;t\in T$ such that
$$
r+s\equiv t\;(\text{mod}\;20)
\qquad
$$
Thus the congruence
$$
3^x+5^z\equiv 4^y\;(\text{mod}\;20)
\;\;\,
$$
has no positive integer solutions, hence the equation
$$
3^x+5^z=4^y
\qquad\qquad\;\;\;
$$
has no positive integer solutions.
Best Answer
How about the Perfect Cuboid, which is still an open problem, to the best of my knowledge.
This problem leads to a system of Diophantine equations:
$$a^2 + b^2 = d^2$$ $$a^2 + c^2 = e^2$$ $$b^2 + c^2 = f^2$$ (i.e. the face diagonals of the cuboid must be integers) and
$$a^2 + b^2 +c^2 = g^2$$ (i.e. the space diagonal must also be an integer.)
It is easily understandable, and at first glance, looks as if it ought to be solvable with a bit of computer searching, but turns out to be a rather harder than it might seem. The interesting (or frustrating) thing here is that it is possible to find solutions for three of the four equations, but no solution is known that satisfies all four.