Proof plan
Prove that $\min(a,b) = 2$.
Prove that , assuming $a=2$, the numerator can actually leave only so many remainders modulo the denominator.
Prove that for large values of $b$, the remainder cannot be zero.
Check small values of $b$ using number-specific modular tricks.
Beginning
I begin with Mastrem's excellent argument. Indeed, note that the expression for the candidate dividend is symmetric in $a,b$. Note that if $a,b$ have the same parity then the numerator will be odd while the denominator will be even. Therefore, they have dissimilar parities, and we assume that $a$ is the even number and $b$ the odd one so that $a \geq 2 , b \geq 3$ (note : any of them being equal to $1$ can be checked separately).
Then, writing $a=2c$, we go modulo $16$ on the numerator. Note that $2^b \geq 8$ so $2^8|a^8 | a^{2^b}$, hence $16 | a^{2^b}$. On the other hand, $b^4 \equiv 1 \pmod{16}$ for any $b$ odd, therefore $4 | 2^a$ implies that $b^{2^a} \equiv 1 \pmod{16}$. All in all, $a^{2^b}+b^{2^a}+11 \equiv 12 \pmod{16}$. In particular, the largest power of $2$ dividing the numerator is $4$.
Thus, the largest power of $2$ dividing the denominator should also be at most $4$, for the ratio to be an integer. However, note that $2^{\min(a,b)}$ divides the denominator. Therefore, we are forced to have $\min(a,b) \leq 2$, which by the conditions imposed forces $a = 2$ and $b \geq 3$ is odd. We assume this from now on.
Remainder restriction
Once we know that $b$ is an odd number and $a =2 \leq b$ , then we ask ourselves when $$
\frac{2^{2^b} + b^4+11}{4+2^{b}}
$$
is an integer i.e. when $1+2^{b-2}$ divides $2^{2^b} + b^4+11$ (since $4$ already does if $b$ is odd and at least $3$, and $lcm(1+2^{b-2},4) = 4+2^b$). We note that $2^{2^b}$ can only leave certain kinds of remainders modulo $1+2^{b-2}$. Indeed, note that if $2^{b-2} \equiv -1 \pmod{2^{b-2}+1}$. Therefore, if $2^b = k(b-2)+l$ for $0 < l \leq b-3$ (note that $l \neq 0$ as $b-2$ is odd), we get that $$
2^{2^b} \equiv 2^{k(b-2)}2^l \equiv (-1)^k2^l \pmod{2^{b-2}+1}
$$
Therefore,
$$
2^{2^b} + b^4+11 \equiv \pm 2^{l} + b^4+11 \pmod{2^{b-2}+1}
$$
for some $l \in \{0,1,...,b-3\}$. The question is, is it possible for the RHS quantity to be equivalent to $0$ modulo $2^{b-2}+1$? For this, some easy estimates are in line.
Note that for large enough $b$, it is true that $2^{b-2}+1 > 2^{b-3}+b^4+11$, therefore for $b-3>l>0$, it is true that $0< 2^l+b^4+11 < 2^{b-2}+1$. On the other hand, note that it is obvious that $-2^{b-3}+b^4+11>-2^{b-2}-1$. Hence, for large enough $b$, it follows that if the expression is equivalent to zero, it literally equals zero i.e. there exists $l \in \{0,1,...,b-3\}$ with $b^4+11 = 2^l$.
So when is $b^4+11 = 2^l$? The answer lies in noting that $b^2 \equiv 1 \pmod{8}$ and therefore $b^4 \equiv 1 \pmod{8}$ for all odd $b$, hence $b^4+11$ cannot be a multiple of $8$ for any odd $b$. Eliminating the small cases, it follows that the equation has no solutions over the integers.
Estimation
Therefore, we are only left to ask, when is $2^{b-2}+1 > 2^{b-3}+b^4+11$? This reduces to $2^{b-3} > b^4+10$, and if one knows their powers of $2$ well (I know I had a dull childhood if I had to memorize powers of $2$), then one sees that $2^{18} = 262144$ and $21^4 < 210000$ so $b=21$ works. One can prove that if the inequality holds for $b$, it does so for $b+1$ as well (by induction). Therefore, there are no solutions for $b \geq 21$ using the logic above.
Finishing the solution
Thus, we are left with $a=2$ and $b$ odd, $b \leq 20$. We can actually do better, though : we have to check if $1+2^{b-2}$ divides $2^{2^b} + b^4+11$, recall. Note that if $b$ is odd, then $1+2^{b-2}$ is actually a multiple of $3$. Therefore, the same should be true of $2^{2^b}+b^4+11$. However, $2^{2^b}+11$ is a multiple of $3$, so it follows that $b^4$ is a multiple of $3$ i.e. $b$ is a multiple of $3$. This leaves only $3,9,15$!
$b=3$ works.
For $b=9$, $1+2^{b-2}$ is a multiple of $43$ by computation, while on the RHS, $9^2 \equiv -5 \pmod{43}$ so $9^4 \equiv 25 \pmod{43}$, and $2^{2^9} \equiv 2^{512} \equiv 2^{8} \equiv 41 \pmod{43}$ (using Euler's theorem with $\phi(43) = 42$), so the RHS modulo $43$ is $41+25+11 \neq 0 \pmod{43}$.
For $b=15$, the LHS $1+2^{b-2} = 8193$ is a multiple of $2731$ (yes, I'm stretching far here, bear with me!) while on the RHS, $2^{2^{15}} = 2^{32768}\equiv 2^8 = 256 \pmod{2731}$, and $15^4 \equiv 1467 \pmod{2731}$,so the RHS modulo $2731$ is $1467 + 256+11 \neq 0$.
It follows that $a=2,b=3$ is the only solution.