Number of solutions to $a^x+b^y=c^z$ with $\gcd(a,b)>1$

We show that there are at most two solutions in positive integers $(x,y,z)$ to the equation $a^x+b^y=c^z$ for positive integers $a$, $b$, and $c$ all greater than one, with just one exceptional case when $\gcd(a,b)=1$, and just one exceptional infinite family of cases when $\gcd(a,b)>1$ (two solutions $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ are considered the same solution if $\{a^{x_1},b^{y_1}\}=\{a^{x_2},b^{y_2}\}$). The case in which $\gcd(a,b)=1$ has been handled in a series of successive results by Scott and Styer, Hu and Le, and Miyazaki and Pink, who showed that there are at most two solutions, excepting $(\{a,b\},c)=(\{3,5\},2)$, which gives three solutions. So, here we treat the case $\gcd(a,b)>1$, showing that in this case there are at most two solutions, excepting $(a,b,c)=(2^u, 2^v,2^w)$ with $\gcd(uv,w)=1$, which gives an infinite number of solutions. This generalizes the work of Bennett, who proved, for both $\gcd(a,b)=1$ and $\gcd(a,b)>1$, there are at most two solutions $(y,z)$ to the equation $a+b^y=c^z$, and conjectured there are exactly eleven $(a,b,c)$ giving two solutions to this equation (assuming $b$ and $c$ are not perfect powers).
For both $\gcd(a,b)=1$ and $\gcd(a,b)>1$, there are an infinite number of $(a,b,c)$ giving two solutions $(x,y,z)$ to the title equation, which are described in detail in this and a cited previous paper.
In a further result, in which we no longer say that two solutions $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ are considered the same solution if $\{a^{x_1},b^{y_1}\}=\{a^{x_2},b^{y_2}\}$, we list all cases with more than two solutions.