Here is a reduction to finitely many cases.

Without loss of generality, $b\geqslant a$. Then we get $$2^a3^a(3^n-2^n)=5^d4^d(5^m-4^m)\tag{1}$$, where $n=b-a\geqslant 0$, $m=c-d$. If $n=0$, we get $m=0$, that's a trivial solution. Further let $n>0$, then $m>0$.

Considering 2-adic valuations of both sides of (1), we get $a=2d$, so (1) reads as $$9^d(3^n-2^n)=5^d(5^m-4^m).\tag{2}$$
Since 5 divides $3^n-2^n$, we conclude that $n$ is even. If $n=2$, looking at 5-adic valuations of both sides of (2) we conclude that $d=1$, then $m=2$, it is a solution. Further let $n>2$. Reducing (2) modulo 3 yields that $m$ is even. Then reducing modulo 8 yields that $d$ is even. Lifting the Exponent Lemma (for a power of 3 which divides $5^m-4^m$) yields that $m$ is divisible by $9^{d-1}$, and (for a power of 5 which divides $3^n-2^n$) that $n$ is divisible by $5^{d-1}$. 
Note that if $n>2m$, then LHS of (2) is $9^d3^n(1-(2/3)^n)\geqslant 9^d3^{2m+1}(1-(2/3)^4)>5^d5^m>$RHS(2), a contradiction. Thus $\min(n,2m)=n$. 
Therefore $2^{n}$ divides $3^{n+2d}-5^{d+m}=(3^{n/2+d}-5^{m/2+d/2})(3^{n/2+d}+5^{m/2+d/2})$. One of these two brackets is not divisible by 4 (and both brackets are non zero), and we conclude that
$$2^{n-1}\leqslant 3^{n/2+d}+5^{m/2+d/2}.\tag{3}$$ 
We have $3^{2d+n}>9^d(3^n-2^n)=5^d(5^m-4^m)=5^{d+m}(1-(4/5)^m)\geqslant \frac{9}{25}5^{d+m}$, therefore $3^{n/2+d}>\frac35 5^{m/2+d/2}$ and (3) yields $$2^{n-1}<\frac83 3^{n/2+d},$$
but since $n/2\geqslant 5^{d-1}$, this has only finitely many solutions in $(n,d)$ which may be all checked by hands.