Find all positive integers $(a,b)$ such that $\displaystyle\frac{a^{2^{b}}+b^{2^{a}}+11}{2^{a}+2^{b}}$ is an integer 
Find all positive integers $(a,b)$ such that
$\displaystyle\frac{a^{2^{b}}+b^{2^{a}}+11}{2^{a}+2^{b}}$ is an
integer.

The machine/code says that $a=2$ and $b=3$ are suitable up to symmetry. And I bet that these are the only solutions when $a≤b$. If $a=0$, then $1+2^b≤b+11$  implies $b≤3$ and the only solution is $b=1$. In a similar way $a=1$ implies  $b=0$. Since the formula is symmetrical, we can assume $a,b>1$.
$a$ and $b$ cannot have the same parity, because the numerator will be odd and the denominator will be even, which is impossible. Suppose $a≥2$ is even and $b≥3$ is odd. The pair $(a,b)=(2,3)$ is a solution giving $\displaystyle\frac{a^{2^{b}}+b^{2^{a}}+11}{2^{a}+2^{b}}=29$. For $a=2$, there is no other solution for $b≤25$. For $a=4$ and $a=6$, there is no solution for $b≤25$.
But I don't think this reasoning will be enough for a complete proof...
 A: As noted by the comments one of $a,b$, say $b$ must be $2$ and the parity of $a$ and $b$ must differ. So let us assume that $a$ is odd.
We first make the following claim:

Claim 1: Let $c$ and $a'$ be nonnegative integers. Then for some nonegative integer $d\le a'$ the equation holds:
$$2^c \equiv_{2^{a'}+1} \pm 2^d.$$

Proof: Induction on $c$. Clearly true for $c \le a'$. Now
$2^{a'+1} \equiv_{2^{a'}+1} -1$ so
$$2^{c} \quad = \quad 2^{c-a'}2^d \quad \equiv_{2^{a'}+1} \quad -1 \times 2^{c-a'} \pmod{(2^{a'}+1)}.$$ However, by the induction hypothesis, there is an integer $d' \le a'$ such that the equation $$2^{c-a'} \equiv_{2^{a'}+1} \pm 2^{d'}$$ holds, so from this Claim 1 follows. $\surd$

Claim 2: For an integer $a>31$ the strict inequality $$a^4+11<2^{a-3}$$ holds.

Proof: Induction on $a$. Check directly for $a=31$, and then note that $(a+1)^4+11 < 2×(a^4+11)$, whereas $2^{(a+1)-3} = 2×2^{a-3}$. So informally, the RHS of the inequality, already larger than LHS, at least doubles when $a$ increases by $1$, whereas the LHS does not increase by such a large factor. $\surd$

We now use this to show there is no solution $(a,b)$ with $a$ odd and $a>31$, and with $b=2$. To this end, with $b=2$ and $a$ odd, the denominator becomes $4 \times 2^{a-2}+1$, and the numerator becomes $a^4+2^{2^a}+11$. Thus, it suffices to show
that $2^{a-2}+1$ cannot divide $11+a^4+2^{2^{a}}$ for $a > 31$.
Now, by the Claim 1, $$2^{2^{a}} \equiv_{2^{a-2}+1} \pm 2^d$$
for some nonegative integer $d \le a-2$. We consider 2 cases:
Case 1: $ \ 2^{2^{a}} \equiv_{2^{a-2}+1} -2^d$ for some nonnegative integer $d$. Then for $2^{a-2}+1$ to divide $11+a^4+2^{2^{a}}$, the equation
$a^4+11-2^d \equiv_{2^{a-2}+1} 0$ must hold.  So from this the following holds:

For there to be an integer $a>31$ such that $2^{a-2}+1$ divides $11+a^4+2^{2^{a}}$ for some nonnegative integer $d$,  there has to be such an $a,d$ for which either the equation (a) $a^4+11-2^d=0$ or the
equation (b) $a^4+11-2^d =n \times (2^{a-2}+1)$ holds.

However, there is no such $a,d$ that satisfies either of these equations (a), (b). [Indeed, the equation $a^4+11-2^d=0$  does not hold for any nonnegative integers $a,d$. [Indeed, checking mod 16, this cannot hold $d \ge 4$. And by exhaustive search this cannot hold for $d \le 4$ either.] And as the
strict inequality $a^4+11< 2^{a-2}$ holds for $a>31$ by Claim 2,  the
equation $a^4+11-2^d =n \times (2^{a-2}+1)$ cannot hold for any nonnegative integer $d$ and any integer $a \ge 31$ either.]
Case 2: $ \ 2^{2^{a}} \equiv_{2^{a-2}+1} + 2^d$ for some nonnegative
integer $d$.  Then we can assume that $d<a-2$ otherwise revert to Case 1. But then $|2^{a-2}-2^d| \ge 2^{a-3}$, so the only way that
$a^4+11+2^d$ is a multiple of $2^{a-2}+1$ is if $a^4+11$ is at least $|2^{a-2}-2^d|$ $\ge$
$2^{a-3}$, and this is impossible for $a \ge 31$ by Claim 2.
So we have indeed shown that there are no solutions $(a,b)$ with $a >31$ and $b=2$. So we can then reasonably finish the original problem simply by checking directly each of the $16$ solutions $(2k+1,2)$; $k=0,1,2,\ldots, 15$ [or we can trust Servaes' calculations in the Comments.]
