Theorem: Odd Numbers not Sum of Prime and Power

The sequence of odd numbers which cannot be expressed as the sum of a perfect power and a prime number begins:
:$1, 5, 1549, 1 \, 771 \, 561, \ldots$
{{OEIS|A119747}}
It is not known if there are any more terms.

Proof:
The cases $1$ and $5$ are trivial.
Now we show that $1549 - a^b$ is never prime for $a \ge 1$ and $b \ge 2$.
It suffices to show the result for prime values of $b$.
We first prove:
{{begin-eqn}}
{{eqn | n = 1
      | l = 2
      | o = \divides
      | r = a
}}
{{eqn | n = 2
      | l = 3
      | o = \divides
      | r = a
      | rr = \text {if } b = 2
}}
{{eqn | n = 3
      | l = a^b
      | o = \not \equiv
      | r = 4
      | rr = \pmod {10}
}}
{{end-eqn}}
For $(1)$:
Suppose $a$ is odd.
Then so is $a^b$.
Therefore $1549 - a^b$ is even.
Hence $1549 - a^b$ is prime only if it is equal to $2$.
But $1547$ is not a perfect power.
{{qed|lemma}}
For $(2)$:
Suppose $b = 2$ and $a$ is not divisible by $3$.
By {{Corollary|Square Modulo 3|3}}:
:$3 \divides \paren {a^2 - 1}$
Hence:
:$3 \divides \paren {1548 - a^2 + 1} = 1549 - a^b$
So $1549 - a^b$ is prime only if it is equal to $3$.
But $1546$ is not a perfect square.
{{qed|lemma}}
For $(3)$:
Suppose $a^b \equiv 4 \pmod {10}$.
Then $1549 - a^b \equiv 5 \pmod {10}$
So $1549 - a^b$ is prime only if it is equal to $5$.
But $1544$ is not a perfect power.
{{qed|lemma}}
Hence we only need to check the cases:
:For $b = 2: \: 6^2, 12^2, 18^2, 24^2, 30^2, 36^2$ since $42^2 > 1549$
:For $b \ne 2: \: 2^3, 2^5, 2^7, 4^3, 4^5, 6^3, 8^3, 10^3$ since $2^{11}, 4^7, 6^5 > 1549$
of which $12^2 \equiv 18^2 \equiv 4^3 \equiv 4^5 \equiv 4 \pmod {10}$, so they are rejected.
We have:
{{begin-eqn}}
{{eqn | l = 1549 - 6^2
      | r = 1513
}}
{{eqn | r = 17 \times 89
}}
{{eqn | l = 1549 - 24^2
      | r = 973
}}
{{eqn | r = 7 \times 139
}}
{{eqn | l = 1549 - 30^2
      | r = 649
}}
{{eqn | r = 11 \times 59
}}
{{eqn | l = 1549 - 36^2
      | r = 253
}}
{{eqn | r = 11 \times 23
}}
{{eqn | l = 1549 - 2^3
      | r = 1541
}}
{{eqn | r = 23 \times 67
}}
{{eqn | l = 1549 - 2^5
      | r = 1517
}}
{{eqn | r = 37 \times 41
}}
{{eqn | l = 1549 - 2^7
      | r = 1421
}}
{{eqn | r = 7^2 \times 29
}}
{{eqn | l = 1549 - 6^3
      | r = 1333
}}
{{eqn | r = 31 \times 43
}}
{{eqn | l = 1549 - 8^3
      | r = 1037
}}
{{eqn | r = 17 \times 61
}}
{{eqn | l = 1549 - 10^3
      | r = 549
}}
{{eqn | r = 3^2 \times 61
}}
{{end-eqn}}
and none of the above are prime.
{{qed|lemma}}
For $11^6$, note the factorizations from Difference of Two Powers:
:$11^6 - m^2 = \paren {11^3 - m} \paren {11^3 + m}$
:$11^6 - n^3 = \paren {11^2 - n} \paren {1 + 11^2 n + 11^4 n^2}$
therefore if $11^6 - a^b$ is a prime, $b$ cannot be divisible by $2$ or $3$, unless $11^3 - m$ or $11^2 - n$ is equal to $1$.
For $m = 11^3 - 1$ and $n = 11^2 - 1$:
:$11^6 - m^2 = 2661 = 3 \times 887$
:$11^6 - n^2 = 1757161 = 7 \times 173 \times 1451$
and both are not primes.
Hence none of the above cases hold.
{{qed|lemma}}
Since $\log_2 \paren {11^6} \approx 20.8$, we must have $b \le 19$.
By the conditions above, $b$ can only be one of $5, 7, 11, 13, 17, 19$.
Moreover $a$ cannot be odd unless $11^6 - a^b = 2$.
However $11^6 - 2 = 1771559$ is prime, and thus not a perfect power.
The cases we need to check are:
:$a^b = 2^5, 6^5, 10^5, 12^5, 14^5, 2^7, 6^7, 2^{11}, 2^{13}, 2^{17}, 2^{19}$
since $18^5, 10^7, 6^{11} > 11^6$, and $4, 8, 16$ are square/cube.
We have:
{{begin-eqn}}
{{eqn | l = 11^6 - 2^5
      | r = 1771529
}}
{{eqn | r = 23 \times 77023
}}
{{eqn | l = 11^6 - 6^5
      | r = 1763785
}}
{{eqn | r = 5 \times 352757
}}
{{eqn | l = 11^6 - 10^5
      | r = 1671561
}}
{{eqn | r = 3^2 \times 79 \times 2351
}}
{{eqn | l = 11^6 - 12^5
      | r = 1522729
}}
{{eqn | r = 13 \times 117133
}}
{{eqn | l = 11^6 - 14^5
      | r = 1233737
}}
{{eqn | r = 311 \times 3967
}}
{{eqn | l = 11^6 - 2^7
      | r = 1771433
}}
{{eqn | r = 31 \times 57143
}}
{{eqn | l = 11^6 - 6^7
      | r = 1491625
}}
{{eqn | r = 5^3 \times 11933
}}
{{eqn | l = 11^6 - 2^{11}
      | r = 1769513
}}
{{eqn | r = 17 \times 104089
}}
{{eqn | l = 11^6 - 2^{13}
      | r = 1763369
}}
{{eqn | r = 41^2 \times 1049
}}
{{eqn | l = 11^6 - 2^{17}
      | r = 1640489
}}
{{eqn | r = 31 \times 52919
}}
{{eqn | l = 11^6 - 2^{19}
      | r = 1247273
}}
{{eqn | r = 17 \times 73369
}}
{{end-eqn}}
and none of the above are prime.
{{qed|lemma}}
{{ProofWanted|In addition to the above, we also need to demonstrate that for $5 < n < 1549$, an odd number does have such an expression.}}