(i) If $a=1,$ then $b=1.$ If $a>1,$ take any $p_{prime} \mid a,$ then $$bv_p \left(a \right) + cv_p \left(b \right) = av_p \left(c \right).$$If $p \nmid b,$ then $$bv_p \left(a \right) = av_p \left(c \right) \Rightarrow p^{v_p \left(a \right)} \mid v_p \left(a \right) \Rightarrow v_p \left(a \right) \geqslant 2^{v_p \left(a \right)} \Leftrightarrow v_p \left(a \right) = 0,$$which is impossible. So $p \mid b.$ (ii) If $b \geqslant a,$ then $a^a \mid a^b \mid c^a \implies a \mid c,$ or $c = ka \ \left(k \geqslant 1 \right).$ Then $$a^b \cdot b^{ka} = \left(ka \right)^a \iff a^{b-a} \cdot b^{ka} = k^a \Rightarrow k^a \geqslant b^{ka} \Leftrightarrow k \geqslant b^{k},$$So $b=1, \ a = 1, \ c = 1.$ (iii) Take any positive integer $x,$ and $a=c=x^x, \ b = x^{x-1}.$

Doxuantrong wrote: (i) If $a=1,$ then $b=1.$ If $a>1,$ take any $p_{prime} \mid a,$ then $$bv_p \left(a \right) + cv_p \left(b \right) = av_p \left(c \right).$$If $p \nmid b,$ then $$bv_p \left(a \right) = av_p \left(c \right) \Rightarrow p^{v_p \left(a \right)} \mid v_p \left(a \right) \Rightarrow v_p \left(a \right) \geqslant 2^{v_p \left(a \right)} \Leftrightarrow v_p \left(a \right) = 0,$$which is impossible. So $p \mid b.$ (ii) If $b \geqslant a,$ then $a^a \mid a^b \mid c^a \implies a \mid c,$ or $c = ka \ \left(k \geqslant 1 \right).$ Then $$a^b \cdot b^{ka} = \left(ka \right)^a \iff a^{b-a} \cdot b^{ka} = k^a \Rightarrow k^a \geqslant b^{ka} \Leftrightarrow k \geqslant b^{k},$$So $b=1, \ a = 1, \ c = 1.$ (iii) Take any positive integer $x,$ and $a=c=x^x, \ b = x^{x-1}.$ Can you please explain the first part, from where did you get the equations?

i) Suppose $p\mid a$ where $p$ is prime. Since $c^a$ is a multiple of $p$, so is $c$, so $p\mid c$. Then if $p^m\Vert a $ and $p^n\Vert c$, write $a=p^mx$ and $c=p^ny$ for some $x,y$ not divisible by $p$. The equation becomes $$ (p^mx)^b\cdot b^{(p^ny)}=(p^ny)^{(p^mx)}$$and if $p\nmid b$, equating the powers of $p$ on both sides gives $mb=np^mx$. But then $p\mid b$, contradiction. So we must have $p\mid b$. ii) If $b\geq a$, then we have $$a^{a+c}=a^a\cdot a^c\leq a^b\cdot b^c=c^a\implies a^{a+c}\leq c^a$$Either $a=c$, $a>c$, or $a<c$. Note that if one of $a,b,c$ is $1$, it's fairly straightforward to show that we must have $a=b=c=1$. If $a=c$, then we obtain $a=c=1$ which results in a solution $(a,b,c)=(1,1,1)$. If $a>c$, then $$a^{2c}<a^{a+c}\leq c^a <a^a$$which is a clear contradiction unless $a=1$. If $c>a$, then we show that $c$ is unbounded unless $c=1$. Note that $$(a^2)^{a}\leq a^{a+c}\leq c^a$$so $a^2\leq c$. Now if $a^n\leq c$ for some integer $n$, then $$a^{(1+a^{n-1})^{a}} \leq a^{a+c}\leq c^a\implies a^{1+a^{n-1}}\leq c$$and $n<1+a^{n-1}$ unless $a=1$. So if $a\neq 1$, starting from $n=2$, we can repeatedly establish larger and larger lower bounds on $c$ to derive a contradiction. So in this case, we also have $a=b=c=1$. In all cases, we conclude that $a=b=c=1$ so the only solution when $b\geq a$ is $a=b=c=1$. iii) To show that infinitely many solutions exist, take the family $(a,b,c)=(t^t,t^{t-1},t^t)$ for any positive integer $t$ which is easily shown to work.

(iii): Construction: $(a,b,c)=(x^x, x^{x-1}, x^x).$ Note, that with this construction, there are infinite, positive integer, values of $x,$ hence, this equation has infinite solutions.