If $11^m\cdot 5^n-3^p\cdot 2^q=1$ where $m,n,p,q$ are non-negative integers,Find all $m,n,p,q$ If $11^m\cdot 5^n-3^p\cdot 2^q=1$ where $m,n,p,q$ are non-negative integers, Find all $m,n,p,q$.
It seems $(1,1,3,1),(0,1,0,2),(0,2,1,3)$ are the only solutions.
Now, the idea is to plug $m=1+x,n=1+y,p=3+z,q=1+r$ with $x,y,z,r \ge 1$
It gives me $55(11^x\cdot 5^y)-54(3^z\cdot 2^r)=1$..How to proceed from here?

A PARTIAL ANSWER.-Seeking positive solutions $mnpq\ne 0$ (or its impossibility) we prove that if $n>1$ then there are no solutions. Besides we give restrictions on the exponents $m,p,q$
►Consider the equation $$11x-3y=1\qquad(*)$$
A solution of $(*)$ is $(x,y)=(5,18)$ so the general solutions is given by
$$(x,y)=(3t+5,11t+18)$$ It follows the equations
$$\begin{cases}3t+5=11^{m-1}5^n\\11t+18=3^{p-1}2^q\end{cases}\qquad(**)$$
It is seen in $(**)$ that $t$ must be multiple of $5$ (first equation)and of $2$ (second equation) hence $t=10s$ so one has 
$$30s+5=11^{m-1}5^n$$
If $n>1$ then $1\equiv 5\pmod {10}$ thus since we are looking for $n$ positive, $\color{red}{n=1}$.
 Furthermore, if $p>1$ we get $t=30s$ so one has 
$$\begin{cases}900s+5=11^{m-1}5^n\\330s+18=3^{p-1}2^q\end{cases}\qquad(***)$$ and the second equation in $(***)$ gives $8\equiv3^{p-1}2^q\pmod {10}$ so 
$$(p-1,q)\in \{(1,4),(2,1),(3,2),(4,3)\}\text{modulo}\space 4$$
The proposed equation becomes in the following 
four possibilities to study
$$\begin{cases}5\cdot11^m-3^{2+4r}2^{4s}=1\\5\cdot11^m-3^{3+4r}2^{1+4s}=1\\5\cdot11^m-3^{4r}2^{2+4s}=1\\5\cdot11^m-3^{1+4r}2^{3+4s}=1\end{cases}$$
