Diophantine equation in primes

Question

Let $p$ and $q$ be primes. Does the equation below has any solutions at all? I can't find any. How would one approach this sort of problem?

$$p-18q = 289$$

Answer 1

The solutions upto $q=1000$ are

? forprime(q=1,1000,if(isprime(18*q+289,2)==1,p=18*q+289;print(p,"   ",q,"   ",i
sprime(p,2),"    ",isprime(q,2))))
379   5   1    1
487   11   1    1
523   13   1    1
631   19   1    1
811   29   1    1
1063   43   1    1
1567   71   1    1
1783   83   1    1
2143   103   1    1
2251   109   1    1
2647   131   1    1
2791   139   1    1
2971   149   1    1
3511   179   1    1
3547   181   1    1
3727   191   1    1
4483   233   1    1
4591   239   1    1
5023   263   1    1
5167   271   1    1
5347   281   1    1
5563   293   1    1
5923   313   1    1
6247   331   1    1
6571   349   1    1
7507   401   1    1
7867   421   1    1
8191   439   1    1
8263   443   1    1
8623   463   1    1
9127   491   1    1
9343   503   1    1
10531   569   1    1
10567   571   1    1
11071   599   1    1
11827   641   1    1
11863   643   1    1
12043   653   1    1
12583   683   1    1
12907   701   1    1
13591   739   1    1
13807   751   1    1
14851   809   1    1
14887   811   1    1
15391   839   1    1
15643   853   1    1
15823   863   1    1
16183   883   1    1
16831   919   1    1
17011   929   1    1
17443   953   1    1
18127   991   1    1
?

To find such solutions, it is best to let run $q$ through the primes and simply verify whether $p:=18q+289$ is prime. You can rule out some modulo-equivalences for $q$ before , but that's it.

If we allow $q$, which are not prime, Dirichlet's theorem ensures that there are infinite many primes $p=18q+289$ because of $gcd(18,289)=1$.

This is probably true if we restrict $q$ to be prime as well, but I do not think that this has been proven.