Assume that $3x+4y$ and $4x+5y$ are both powers of $7$. Then $(3x+4y)+(4x+5y)=7x+9y$ is divisible by $7$. It means that $y$ is divisible by 7, and then it is easy to conclude that $x$ is also divisible by 7. This implies that $x=7x_1$ and $y=7y_1$. As a result $(3x_1+5y_1)(4x_1+5y_1)=7^{z-4}$. Using infinite descent we get there is no solution in this case.
If $3x+4y$ and $4x+5y$ are not both powers of $7$, then one of them is $1$ and it is easy to check that it is impossible.
Hence no solution