Q:

How to prove there are no solutions for $5^{3+a} = 11^{2+b} + 4$ except $a=b=0$? 

Working out some similar questions with my systematic approach as done for instance in this older question (and in some other questions) I could not yet find the equivalent proof of nonexistence of a solution in $a,b \gt 0$ 
$$ 5^{3+a} = 11^{2+b} + 4 \tag 1$$ 
which I usually rewrite as
$$ {5^a -1 \over 11^2} = {11^b - 1\over 5^3 } \tag 2$$ 
Possibly there is a simpler method than mine by some cleverer modular considerations.            
Well, I didn't check whether we can deduce from the fact that $4=2^2$ being a square number there is only a finite (already known) list of possible solutions for $x^a-y^b = z^2$ for $a,b>2$ which would solve this immediately.  However, I'd prefer the other way: 

I'd like to see a proof using modularity to understand, why or where my own method fails or why it might need only more/excessive effort.  

A:

I solved the problem myself - I'd just overlooked one test.  Here it is, how it went. 

The ansatz is to determine the required structures in the exponents $a$ and $b$ to allow compatible sets of primefactors in the lhs and rhs:
$$ {5^a-1 \over 11^2 } = {11^b-1 \over 5^3 } \tag 1$$
Initialization
To have exactly $11^2$ as factor in the numerator of the lhs we need that $a=5 \cdot 11 \cdot a_2$ : 
$$ \begin{array}{} 
   \text{lhs :}
 & \large {5^a-1 \over 11^2 } 
 &=\large {5^{55 \cdot a_2}-1 \over 11^2 }
 & \underset{a_2=1}= 
 & 2^2.71.103511.\text{<big>}
\end{array}$$
To have exactly $5^3$ as factor in the numerator of the rhs we need that $b=5^2  \cdot b_2$ 
$$ \begin{array}{}
 \text{rhs :} 
& \large {11^b-1 \over 5^3 } 
&=\large  {11^{25 \cdot b_2}-1 \over 5^3 } 
& \underset{b_2=1}= 
& 2.3001.3221.\text{<big>}
\end{array}$$
Step 1 : primefactor $2$
Next, we adapt the exponents so that lhs and rhs have the primefactor $2$ to the same exponent.  We need $a_2$ and $b_2$ being even, so we define $a_2=2 a_3$ and $b_2 = 2 b_3$ finding
$$ \begin{array}{} 
   \text{lhs :}
 & \large {5^{55\cdot 2 \cdot a_3}-1 \over 11^2 }
 & \underset{a_3=1}= 
 & 2^3.3.23.67.71.521.5281.\text{<big>} \\
 \text{rhs :} 
& \large  {11^{25 \cdot 2 \cdot b_3}-1 \over 5^3 } 
& \underset{b_3=1}= 
& 2^3.3.3001.3221.13421.\text{<big>}
\end{array}$$
Step 2 : primefactor $3001$
Next, we adapt the exponent in the lhs that it has primefactor $3001$. We need the exponent containing $250$ so $a_3$ needs to provide the factor $5^2$ so we define $a_3= 5^2 a_4$ finding
$$ \begin{array}{} 
   \text{lhs :}
 & \large {5^{55 \cdot 2 \cdot 5^2 \cdot a_4}-1 \over 11^2 }
 & \underset{a_4=1}= 
 & 2^3.3.23.67.71.101.251.401.521.1901.3001.\text{<big>} 
\end{array}$$
Step 3 : primefactor $251$
Finally, we adapt the exponent in the rhs that it has primefactor $251$. We need thus the exponent containing $250$ as on the lhs, so $b_3$ needs to provide the factor $5$ so we define $b_3= 5 b_4$ finding
$$ \begin{array}{} 
 \text{rhs :} 
& \large  {11^{25 \cdot 2 \cdot 5 \cdot b_4}-1 \over 5^3 } 
& \underset{b_4=1}= 
& 2^3.3.5.251.3001.3221.13421.\text{<big>}
\end{array}$$
Proof by contradiction on primefactor $5$
Here in the primefactorization of the rhs we find an additional factor $5$ which cannot be compensated by varying the exponent in the lhs.  
We needed only partial adaptions of exponents (but of course all steps are mandatory), and still arrived at the expected contradiction:      

We cannot have exponents $a,b>0$ providing equality in (1) and thus the only solution is the initial one $ 5^3 = 11^2 + 4$

What we have explicitely used were the primefactors $2,3001,251$, and likely direct considerations on modularities (as usual with such problems) should use that moduluses (possibly one does not need the primefactor $2$).  However this does not exclude that my basic approach could uncover more/alternative solutions for the disproof.               
Appendix: a provisorical Pari/GP routine for solving such problems systematically can be obtained if desired
For an extended explanation look at this older MSE-answer or this

