Finding solutions to the diophantine equation $7^a=3^b+100$

Fair to say that creating this proof (especially predicting that the ordered pair of primes $811$ and $3889$ would work) is out of the reach of hand computations, although everything used can be confirmed by hand. We have $7^a = 3^b + 100,$ and suspect that the largest solution is $343 = 243 + 100.$ Well, take $7^a - 343 = 3^b - 243.$ This becomes $$ 343 ( 7^x - 1) = 243 ( 3^y - 1). $$ We are going to prove that we cannot accomplish this with $x,y \geq 1.$

Assuming $x,y \geq 1:$ Since $$ 7^x \equiv 1 \pmod {243}, $$ we find $$ 81 | x \Longrightarrow 27 | x. $$

$$ 7^{27} - 1 = 2 \cdot 3^4 \cdot 19 \cdot 37 \cdot 109 \cdot 811 \cdot 1063 \cdot 2377 \cdot 2583253 $$ This divides $7^x - 1.$ In particular, $811 | (7^x - 1),$ and so $811 | (3^y - 1.)$

jagy@phobeusjunior:~$ ./order 3 811
811   810 = 2 * 3^4 * 5

$$ 3^y \equiv 1 \pmod {811} \Longrightarrow 810 | y \Longrightarrow 81 | y. $$

$$ 3^{81} - 1 = 2 \cdot 13 \cdot 109 \cdot 433 \cdot 757 \cdot 3889 \cdot 8209 \cdot \mbox{BIG} $$ In particular, $3^{81} - 1$ is divisible by $3889,$ so $3^y - 1$ is divisible by $3889.$ In turn, this means that $7^x - 1$ is divisible by $3889.$

$$ 7^x \equiv 1 \pmod {3889} \Longrightarrow 1944 | x \Longrightarrow 243 | x. $$

jagy@phobeusjunior:~$ ./order 7 3889
3889  1944 = 2^3 * 3^5

We have shown $243 | x.$ However, $$ 7^{243} -1 = 2 \cdot 3^6 \cdot 19 \cdot 37 \cdot \mbox{Many More}$$ This means that $$ 729 | (7^x - 1) $$ This contradicts $$ 343 ( 7^x - 1) = 243 ( 3^y - 1) $$ with $x,y \geq 1.$

I learned this technique from Exponential Diophantine equation $7^y + 2 = 3^x$ I also placed three different examples as answers at Elementary solution of exponential Diophantine equation $2^x - 3^y = 7$.

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.