<s>[INST] Finding solutions to the diophantine equation $7^a=3^b+100$ 
Find the positive integer solutions of the diophantine equation $$7^a-3^b=100.$$

So far, I only found this group $7^3-3^5=100$.
 [/INST] note: this answer has been found to be -at least- incomplete, see comments of Gottfried Helms and piquito       
$100$ has no primitive root therefore $7^m$ and $3^n$ are congruent with $1$ modulo $100$ for some integers $m,n$ smaller than $100$ and these numbers should  divide $100$.
We have $7^4\equiv 3^{20}\equiv 1\pmod{100}\Rightarrow 7^{4m}\equiv 3^{20n}\equiv 1\pmod {100}$ which could be a starting point to calculate possible solutions. However we notice in this search that (in the ring $\Bbb Z/100\Bbb Z$ for short) the solution $7^3=3^5+100$ so we have
$$\begin{cases}7^a=3^b+100\\7^3=3^5+100\end{cases}\Rightarrow 7^a-7^3=3^b-3^5$$
For the values $a=1,2,3$ and $b=1,2,3,4,5$ it is not verified except for $(a,b)=(3,5)$ which does not provide another solution. Hence $a\gt 3$ and $b\gt 5$. It follows
$$7^3(7^{a-3}-1)=3^5(3^{b-5}-1)\Rightarrow 7^{a-3}=244=2^2\cdot61\text{ and } 3^{b-5}=344=2^3\cdot43$$ which is absurde. Consequently $(a,b)=(3,5)$ is the only solution.
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