Is $100$ the only square number of the form $a^b+b^a$? Conjecture: $100$ is the only square number of the form $a^b+b^a$ for integers $b>a>1$.
In other words, $(a,b)=(2,6)$ is the only solution. Can we prove/disprove this?
Observations:

*

*The solution mentioned should not come as a surprise, since $2^6+6^2=8^2+6^2=10^2$ is a (non-primitive) Pythagorean triple. It is possible to show that $2^b+b^2$ has no other solutions. See Remark 1.


*In the general case where $a$ is a power of $2$; that is, $a=2^d$ for some positive integer $d$, a similar approach can be followed. See Remark 2.


*We can eliminate some values of $b$ when $a=5^r,6^r$, since no matter the value of $r$, we have $a\equiv5,6\pmod{10}$ respectively.


*PARI/GP code: If the conjecture is true it should only ever print 2 6.
sqfun(a,b)={for(i=2,a,for(j=2,b,if(issquare(i^j+j^i)==1,print(i," ",j))));}


Remark 1: Suppose that there is a positive integer $b$ that admits $2^b+b^2=t^2$ for some integer $t$. Then we can write the equation as $$2^b=(t+b)(t-b)\implies\begin{cases}t+b=2^c\\t-b=2^{b-c}\end{cases}$$ for some positive integer $c>\dfrac b2$. Subtracting the two equations yields $$2b=2^c-2^{b-c}\implies b=2^{b-c-1}(2^{2c-b}-1).$$ If $b$ is odd, it cannot have a factor of $2$, forcing $b-c-1=0\implies t=b+2$ and substituting gives $2^b+b^2=(b+2)^2$, or $2^b=4(b+1)$. No solutions exist.
If $b$ is even, then $b=2k$ for some positive integer $k$, so we must have $$\begin{cases}2^k=s(m^2-n^2)\\2k=2mns\end{cases}$$ for some integers $m,n,s$, so that $2^{mns}=s(m^2-n^2)$. Without loss of generality, let $m>n>0$ and $s>0$. [Servaes: If $mns\ge4$ then $2^{mns}\geq(mns)^2\geq sm^2> s(m^2-n^2)$, so the only solutions with even $b$ are $b=4,6$, and the first case does not yield a square.]
Remark 2: If $b$ is odd, it boils down to the equation $$2^{db}=4\left(b^{2^{d-1}}+1\right)\implies 2^{db-2}-1=b^{2^{d-1}}.$$ [Haran: For $db-2>1$, the LHS is congruent to $3\pmod4$, and since the RHS is a square for $d>1$, we reach a contradiction unless \begin{cases}d=1\implies a=2\quad\text{case covered above}\\db-2=1\implies1=b^{2^{d-1}}\implies 2^{d-1}=0\end{cases} which is impossible.]
If $b$ is even, then $b=2k$ for some positive integer $k$, and the Pythagorean triplet forces $$\begin{cases}2^{dk}=s(m^2-n^2)\\(2k)^{2^{d-1}}=2mns.\end{cases}$$ [Servaes: From the first equation, all three factors on the RHS are powers of two, so $$\begin{cases}m+n=2^u\\m-n=2^v\end{cases}\implies\begin{cases}m=2^{v-1}(2^{u-v}+1)\\n=2^{v-1}(2^{u-v}-1)\end{cases}$$ with $u>v>0$. Since $m$ and $n$ are coprime, we have $v=1$. Plugging this into the first equation yields $$2^{dk}=s(m-n)(m+n)=2^{u+1}s\implies s=2^{dk-u-1}.$$ Substituting this into the second equation yields $$(2k)^{2^{d-1}}=2mns=2(2^{u-1}+1)(2^{u-1}-1)s=(2^{2u-2}-1)2^{dk-u}$$ which is impossible; if we let $k=2^w\ell$ with $\ell$ odd then this implies $\ell^{2^{d-1}}=2^{2u-2}-1$ which by Catalan's conjecture/Mihailescu's theorem is impossible if $d>1$. Note that $u>v$ hence $u\geq2$.]

$\newcommand{\eps}{\varepsilon}$
$\newcommand{\rad}{\mathrm{rad}}$
At least, under the abc conjecture, there can be only finitely many pairs $(a,b)$ with $b>a>1$ coprime such that $a^b+b^a$ is a square.
As a reminder, the conjecture says that to any $\eps>0$ there corresponds some $K_\eps>0$ such that whenever $u,v$, and $w$ are coprime positive integers with $u+v=w$, one has $\rad(uvw)>K_\eps w^{1-\eps}$. Here $\rad(z)$ is the product of all primes dividing $z$ (thus, for instance, $\rad(8)=2$, $\rad(9)=3$, $\rad(10)=10$, $\rad(11)=11$, and $\rad(12)=6$).
Suppose now that $a^b+b^a=c^2$ with coprime integers $b>a\ge 3$ and $c>0$ (the case $a=2$ is resolved above). Applying the abc conjecture with $u=a^b$, $v=b^a$, $w=c^2$, and $\eps=0.05$, and making the key observation $\rad(a^bb^ac^2)\le abc$, we conclude that
  $$ Kc^{2\cdot 0.95} < abc $$
with an absolute constant $K>0$. At the same time, we have $c^2>a^b$ and $c^2>b^a$, implying $a<c^{2/b}$ and $b<c^{2/a}$, respectively. Consequently,
  $$ Kc^{1.9} < c^{(2/b)+(2/a)+1}, $$
showing that either $\frac1b+\frac1a>0.4$, or $Kc^{0.1}<1$. Clearly, there are only finitely many values of $c$ satisfying the latter condition, and to each value corresponds finitely many pairs $(a,b)$. On the other hand, since $\frac1b+\frac13\ge\frac1b+\frac1a>0.4$ implies $b<15$, there are  only finitely many pairs $(a,b)$ satisfying the former condition. Thus, the total number of exceptional pairs $(a,b)$ is also finite.
