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 $7^3-3^5=100$.
 A: 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.
A: You're solving $7^a - 3^b = 100$ in positive integers.
This full solution uses the recent (found in 1998) results by J. Gebel, A. Pethö, G. H. Zimmer, Mordell, etc. (see this paper).
See this paper for some information about Mordell's equations.
mod $7$ gives $b=6k+5$, $k\ge 0$, $k\in\mathbb Z$.
mod $9$ gives $a=3m$, $m\ge 0$, $m\in\mathbb Z$.
$$\left(3\cdot 7^m\right)^3 = 2700 + \left(3^{3k+4}\right)^2$$
http://tnt.math.se.tmu.ac.jp/simath/MORDELL/
shows that $3\cdot 7^m=21$, $3^{3k+4}=\pm 81$, i.e. $(m,k)=(1,0)$, i.e. $(a,b)=(3,5)$.
A: For $\quad(-3000\le a \le 3000)\quad$ there is only one integer solution for $b$.
$$7^a=3^b+100\implies b=\frac{\log(7^a-100)}{\log(3)}\implies (a,b)=(3,5)$$
