(21-03-2020) Update: As no unconditional answer has yet been given, I will include a few more conditions that any solution must satisfy. The results used are rather advanced, so I will only include references, not proofs.
Lemma 5: Let $a$ and $b$ be positive integers such that $a^b+b^a$ is a perfect square. Then $\gcd(a,b)\leq3$.
Proof. Such a pair of positive integers yields a nontrivial integral solution to
$$x^d+y^d=z^2,\tag{3}$$
where $d:=\gcd(a,b)$. This implies $d\leq3$ by this paper${}^1$.$\qquad\square$
Proposition 6: Let $a$ and $b$ be positive integers such that $a^b+b^a$ is a perfect square. Then $\gcd(a,b)\leq2$.
Proof. By the preceding lemma it suffices to show that $\gcd(a,b)=3$ is impossible. If $\gcd(a,b)=3$ then $a^{\tfrac b3}$ and $b^{\tfrac a3}$ are integers satisfying
$$\Big(a^{\tfrac b3}\Big)^3+\Big(b^{\tfrac a3}\Big)^3=z^2,$$
for some integer $z$, which means that, after swapping $a$ and $b$ if necessary, either
$$a^{\tfrac b3}=\frac{x(x^3-8y^3)}{w^2}z^2
\qquad\text{ and }\qquad
b^{\tfrac a3}=\frac{4y(x^3+y^3)}{w^2}z^2,$$
for integers $x$, $y$ and $z$ with $x$ odd and $x$ and $y$ coprime, and $w:=\gcd(3,x+y)$, or
$$a^{\tfrac b3}=\frac{x^4+6x^2y^2-3y^4}{w^2}z^2
\qquad\text{ and }\qquad
b^{\tfrac a3}=\frac{3y^4+6x^2y^2-x^4}{w^2}z^2,$$
for $x$, $y$ and $z$ with $x$ and $y$ coprime and $x$ coprime to $3$, and $w=\gcd(2,x+1,y+1)$. These complete parametrizations of the integral solutions to $(3)$ when $d=3$ are taken from section 7.2 of this article${}^2$.
For the second parametrization, because $x$ is coprime to $3$ we see that
$$\nu_3\Big(a^{\tfrac b3}\Big)=\nu_3(z^2)
\qquad\text{ and }\qquad
\nu_3\Big(b^{\tfrac a3}\Big)=\nu_3(z^2),$$
where $\nu_p(t)$ denotes largest integer $k$ such that $p^k$ divides $t$. It follows that $a\nu_3(b)=b\nu_3(a)$, and from $\gcd(a,b)=3$ it follows that either $\nu_3(a)=1$ or $\nu_3(b)=1$, so either $a$ divides $b$ or $b$ divides $a$, respectively. This means either $a=3$ or $b=3$.
If $a=3$ then the identity
$$3^{\tfrac b3}=a^{\tfrac b3}=\frac{x^4+6x^2y^2-3y^4}{w^2}z^2,$$
shows that $z^2=3^{\tfrac b3}$. The parametrization for $b^{\tfrac a3}$ then shows that $3^{\tfrac b3}$ divides $b^{\tfrac a3}=b$, which quickly implies $b=3$. But $3^3+3^3=54$ is not a perfect square; a contradiction. If $b=3$ a similar argument shows that then $a=3$, and this shows that the second parametrization yields no solutions to our original problem.
For the first parametrization, note that $b$ is even and hence $a^{\tfrac b3}$ is a perfect square, hence so is
$$w^2z^{-2}a^{\tfrac b3}=x(x^3-8y^3)=x^4-8xy^3.$$
This means there is some integer $c$ such that $x^4-8xy^3=(x^2-2c)^2$ and so
$$cx^2+2y^3x-c^2=0.$$
In particular the discriminant $\Delta$ of this quadratic polynomial in $x$ is a perfect square, say $\Delta=(2e)^2$, where
$$4e^2=\Delta=(2y^3)^2-4c(-c)^2=4(y^6+c^3).$$
It is a classical result that then for some nonnegative integer $k$ we have
$$(|e|,c,|y|)=(3k^3,2k^2,k).$$
Plugging this in and solving the quadratic equation for $x$ yields $x\in\{\pm k,\pm2k\}$ where $y=\pm k$. Because $x$ and $y$ are coprime it follows that $k=1$, so $y=\pm1$ and $x\in\{\pm1,\pm2\}$. Then for $a^{\tfrac b3}$ to be a perfect square we must have $\{x,y\}=\{1,-1\}$, but then $b=0$, a contradiction. This shows that the second parametrization also yields no solutions to our original problem. In conclusion $\gcd(a,b)=3$ is impossible. $\quad\square$.
A small step towards a complete proof of the original problem would be to show that any solution other than $\{a,b\}=\{2,6\}$ must have $a$ and $b$ coprime, which seems likely. So for now, I propose the following conjecture:
Conjecture 7: Let $a$ and $b$ be positive integers such that $a^b+b^a$ is a perfect square and $\gcd(a,b)=2$. Then $\{a,b\}=\{2,6\}$.
Of course such solutions yield Pythagorean triples, for which the parametrization is well known. Perhaps arguments similar to those of Proposition 6 can be used here. I hope to give another update resolving this conjecture soon.
References
- H. Darmon and L. Merel, Winding quotients and some variants of Fermat’s last theorem, J.Reine Angew. Math. 490 (1997), 81–100.
- H. Darmon, A. Granville, On the equations $x^p+y^q=z^r$ and $z^m=f(x, y)$, Bulletin of the London Math. Society, no 129, 27 part 6, November 1995, pp. 513–544.
Original answer:
I'll collect a few partial results here. Let $a$, $b$ and $c$ be positive integers with $a,b>1$ such that
$$a^b+b^a=c^2,$$
and let $d=\gcd(a,b)$. First two lemmas that are repeatedly useful.
Lemma 1: If $m$ and $n$ are positive integers with $m>n$ and not both even, such that $m+n$ and $m-n$ are both powers of $2$, then $m=2^k+1$ and $n=2^k-1$ for some positive integer $k$.
Proof. If $m+n=2^u$ and $m-n=2^v$ then
$$m=\frac{(m+n)+(m-n)}2=\frac{2^u+2^v}2=2^{v-1}(2^{u-v}+1),$$
$$n=\frac{(m+n)-(m-n)}2=\frac{2^u-2^v}2=2^{v-1}(2^{u-v}-1),$$
and hence $v=1$ because one of $m$ and $n$ is odd. Then $k=u-v$.$\qquad\square$
Lemma 2: The only perfect power that is one less than a square is $8$.
Proof. There are fairly elementary proofs, but it also follows from Mihailescu’s theorem.$\qquad\square$
Proposition 3: If $a$ is a power of $2$ then $(a,b)=(2,6)$.
Most of this was proved in the original question by TheSimpliFire and Haran.
Proof. Let $a=2^e$. If $b$ is odd then writing
$$(c-b^{2^{e-1}})(c+b^{2^{e-1}})=c^2-b^a=a^b=2^{be},$$
shows that both factors on the left hand side are powers of $2$. Then by Lemma 1 we have $c=2^v+1$ and
$$b^{2^{e-1}}=2^v-1,$$
for some positive integer $v$ because $b$ is odd. Hence by Lemma 2 either $v=1$ or $2^{e-1}=1$. Clearly $v=1$ is impossible, so $2^{e-1}=1$ and so $e=1$. Then comparing exponents shows that $b=v+2$ and so
$$v+2=b=2^v-1,$$
which is easily seen to have no integral solutions. Hence $b$ is even, say $b=2f$. Then we have the following Pythagorean triple:
$$c^2=a^b+b^a=(2^e)^{2f}+(2f)^{2^e}=(2^{ef})^2+((2f)^{2^{e-1}})^2.$$
Then there exist positive integers $k$, $m$ and $n$ with $m>n$ and $\gcd(m,n)=1$ such that either
$$c=k(m^2+n^2),\qquad2^{ef}=k(m^2-n^2),\qquad (2f)^{2^{e-1}}=2kmn,\tag{1}$$
$$\text{or}$$
$$c=k(m^2+n^2),\qquad2^{ef}=2kmn,\qquad (2f)^{2^{e-1}}=k(m^2-n^2).\tag{2}$$
In case the triple is of the form $(2)$, the middle identity shows that $k$, $m$ and $n$ are all powers of $2$, so in particular $n=1$ because $m$ and $n$ are coprime and $m>n$. Then the latter identity shows that
$$(2f)^{2^{e-1}}=k(m^2-1)=k(m-1)(m+1),$$
where the factors $m-1$ and $m+1$ are odd and $k$ is a power of $2$, so both $m-1$ and $m+1$ are $2^{e-1}$-th powers. But for $e>1$ no two $2^{e-1}$-th powers of positive numbers differ by $2$, so $e=1$. Writing $k=2^u$ and $m=2^v$ we see that $u+v+1=f$, where $v\geq1$ because $m>n$.
By comparing powers in the above we find that
$$u+v+1=2^{u-1}(2^{2v}-1)=2^{u-1}(2^v-1)(2^v+1).$$
Of course $2^v+1>2$, so $2^{u-1}=1$ as otherwise
$$2^{u-1}(2^v+1)>2^{u-1}+2^v+1\geq u+v+1,$$
a contradiction. Hence $u=1$ and $2^{2v}-1=v+2$, so also $v=1$. This yields the solution $(a,b)=(2,6)$.
On the other hand, if the Pythagorean triple is of the form $(1)$ then $k$, $m-n$ and $m+n$ are powers of $2$ because
$$2^{ef}=k(m^2-n^2)=k(m-n)(m+n).$$
Because $m$ and $n$ are not both even, by Lemma 1 there exists a positive integer $v$ such that $m=2^v+1$ and $n=2^v-1$, and so the above shows that $k=2^{ef-v-2}$. Plugging this into the other equation yields
$$(2f)^{2^{e-1}}=2kmn=2^{ef-v-1}(2^{2v}-1),$$
and writing $f=2^wg$ with $g$ odd then implies
$$g^{2^{e-1}}=2^{2v}-1,$$
which by Lemma 2 implies that $2^{e-1}=1$, and hence $e=1$. Then $f=kmn$ and if $kmn\geq4$ then
$$2^{kmn}\geq(kmn)^2\geq km^2>k(m^2-n^2),$$
so $f=kmn\leq3$. Then $b\leq6$ and clearly $b=2$ and $b=4$ do not yield solutions.$\qquad\square$
Proposition 4: If $d$ is even then $d=2$.
Proof. Suppose $d=2e$ and let $a=2eA$ and $b=2eB$. Then $e$ is odd as otherwise $c^2$ is the sum of two fourth powers, which is well known to be impossible by a classical result by Fermat. Now
$$c^2=a^b+b^a=(a^{eB})^2+(b^{eA})^2,$$
is a pythagorean triple and hence there exist positive integers $k$, $m$ and $n$ with $m>n$ and $\gcd(m,n)=1$ such that
$$a^{eB}=k(m^2-n^2),\qquad b^{eA}=2kmn,\qquad c=k(m^2+n^2),$$
where we can exchange the roles of $a$ and $b$ if necessary. It is clear that
$$k=\gcd(a^{eB},b^{eA})=\gcd((dA)^B,(dB)^A)^e=d^{e\ell},$$
where $\ell\geq\min\{A,B\}$. In particular $k$ is an $e$-th power, and hence the factorizations
$$(a^B)^e=k(m^2-n^2)=k(m-n)(m+n)
\qquad\text{ and }\qquad
(b^A)^e=2kmn,$$
show that, up to powers of $2$, the factors $m$, $n$, $m+n$ and $m-n$ are also $e$-th powers. That is to say,
$$m=2^tp^e,\qquad n=2^uq^e,\qquad m+n=2^vr^e,\qquad m-n=2^ws^e,$$
for odd positive integers $p$, $q$, $r$ and $s$, and nonnegative integers $t$, $u$, $v$ and $w$, and $t+u+1\equiv v+w\equiv0\pmod{e}$. Then
$$m=\frac{(m+n)+(m-n)}{2}=\frac{2^vr^e+2^ws^e}{2}=2^{v-1}r^e+2^{w-1}s^e,$$
$$n=\frac{(m+n)-(m-n)}{2}=\frac{2^vr^e-2^ws^e}{2}=2^{v-1}r^e-2^{w-1}s^e,$$
and at least one of $m$ and $n$ is odd, so either $v=1$ or $w=1$ (but not both) or $v=w=0$.
If either $v=1$ or $w=1$ (but not both) then $m$ and $n$ are both odd, so $t=u=0$ and hence $e=1$.
If $v=w=0$ then still either $t=0$ or $u=0$ because $m$ or $n$ is odd. If $u=0$ then $e\mid t+1$ and
$$2^{t+1}p^e=2m=(m+n)+(m-n)=r^e+s^e,$$
and so it follows from Fermats last theorem that $e=1$. The same holds if $t=0$.$\qquad\square$
