In fact, there are many more such examples, though very few of them involve twin primes.
With the terminology in the reply above, up to a<10^{323}, there appears to be only one such (a,b,add=2): 794018604377235322848433897872605582794018604377235322848433897872605583 = 891077215721081784886888257701070829 × 891077215721081784886888257701070827.
Using 40 Miller-Rabin iterations, I have obtained the following equalities, all with a LHS involving up to 644 digits:
25187118108481597700189092548093595916723774714848814417444367033247162518711810848159770018909254809359591672377471484881441744436703324717 = 5018676928083894672666012088036109843105301546773725102790665815794441 × 5018676928083894672666012088036109843105301546773725102790665815794437 (shown above)
101936573393229113746896078599158434015231091149521033461905493584141767240967286279147302756939077798218255227325457025854937718962600442639675514396101936573393229113746896078599158434015231091149521033461905493584141767240967286279147302756939077798218255227325457025854937718962600442639675514397 = 319275074807334613749250748026616288651767499918792546380748503672549086703203277071612100663412425258886294180722873606632627934423491347284352444051 × 319275074807334613749250748026616288651767499918792546380748503672549086703203277071612100663412425258886294180722873606632627934423491347284352444047
93919382143599443874360974179992284882076026690881005214561418008150790494258654795862966202627730222978744961557082451286192284357751247078089025462416769391938214359944387436097417999228488207602669088100521456141800815079049425865479586296620262773022297874496155708245128619228435775124707808902546241677 = 9691201274537612443672374967144909095364760038874719917817704809652025255645361670391758426040026736596675168244667860182899459759546967572708946339679961 × 9691201274537612443672374967144909095364760038874719917817704809652025255645361670391758426040026736596675168244667860182899459759546967572708946339679957
981718780415872785585564201031992816866441710921031298607709720433328082774757576994927683441599794030113676908258383896762642955368235301572285359675451074772206407951599583811274848830644325337802597091580076981718780415872785585564201031992816866441710921031298607709720433328082774757576994927683441599794030113676908258383896762642955368235301572285359675451074772206407951599583811274848830644325337802597091580077 = 990817228562297853620579821430841029476907621168021829447531348075063345099849542958089591111031777996054793781667515002540761697546809287372732881673066506036575450357354487060783218299272638998714542277310011 × 990817228562297853620579821430841029476907621168021829447531348075063345099849542958089591111031777996054793781667515002540761697546809287372732881673066506036575450357354487060783218299272638998714542277310007
et cetera.
The StackExchange post A 2021 problem: 20∼21 and 43×47 suggests a heuristic that the number of expected solutions with 2n digits is O(1/n). Judging by the sheer number of solutions, this heuristic is probably incorrect. An alternative model may have to be created.
In approaching this problem I formulated it through expressing it as a~(a+1)=p(p+2k), where p+2k is the prime after p. Simplification yields 100...01a+1=(p+k)^2-k^2, so 100...01a+1+k^2 becomes a square. The problem thus amounts to showing that 1+k^2 is a quadratic residue modulo 10^n+1, and when complete, finding all modular square roots of 1+k^2 and checking that the primality requirements are satisfied. (Namely, that p and p+2k are prime and all numbers in between are composite.)
One clear pattern found is that there are many such pairs with the same number of digits (n). For instance, with k=2 (cousin primes) there are 27 solutions with n=210 and none with 210<n<270. This is something that the heuristic mentioned cannot explain, but it is readily explained with this implementation. For n=210, 10^{210}+1 has 22 distinct prime factors, all of them having 5 as a Q.R.. This means that 5 has 2^{22}\approx 4\times 10^6 square roots mod 10^{210}+1. The density of primes at around 10^{210} is approximately 1 out of 210\ln10\approx 480, so the approximate expected number of solutions satisfying the two primality conditions is \frac{2^{22}}{480^2}\approx 18. Of course, this is also a heuristic as the true fluctuations of primes cannot be accounted for.
Another noteworthy pattern is that there are far more such pairs with k=2 (prime gap 4) than for any other choice of k. Among the 4 solutions Ed Pegg has found, 2 of them have prime gap 4, and one has prime gap 2. As it turns out, for n<323 there is only 1 solution with prime gap 2 but 40 solutions with prime gap 4. This may be related to our use of base 10, as prime factors of 10^n+1 tend to readily be congruent to 1 or 9 mod 10, so 5 is a QR mod 10^n+1 for many n. The same might not hold true for other k, as it is unlikely that all 22 prime factors of 10^{210}+1 should be such that an arbitrary number is a quadratic residue modulo these primes. (when this happens, which is exceedingly rare, there may be several solutions.
Overall, this is an extremely fascinating and deep problem. May 2021 be a uniquely good year!
Extra info: I have obtained that for k=38, 1+k^2=1445=5\times 17^2, so that 1445 is a Q.R. mod 10^n+1 if 5 is one. Therefore, one can observe that for k=38 or 682, it is expected that there would be multiple solutions with n=210 or 270, where there were many solutions for k=2. (For k=682, there is a chance that the numbers between the two primes may not be prime, but considering that heuristically reduces the number by only a factor of 4.)
[1]: https://math.stackexchange.com/questions/3961535/a-2021-problem-20-sim-21-and-43-times-47?fbclid=IwAR1MVuIK6La4FuIkguNz8hk4mXOlA2aOclXLOXr_5yZLC3p_-PxircNXbU8 “stackexchange post”
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