Let $k>1$ be a positive integer, and let $a$, $b$ and $c$ be positive integers such that
$$a^{2k}+b^2=c^{2k}.\tag{0}$$
Let $d:=\gcd(a,c)$ so that $d^{2n}$ divides $a^{2k}$ and $c^{2k}$, and hence also $b^2$. Then $d^k$ divides $b$ and so $(a,b,c)=(dA,d^kB,dC)$ for some positive integers $A$, $B$ and $C$ with $\gcd(A,C)=1$. It follows that also $\gcd(A,B)=\gcd(B,C)=1$, and it is easily verified that
$$A^{2k}+B^2=C^{2k}.$$
So without loss of generality $a$, $b$ and $c$ are pairwise coprime. Then $(a^k,b,c^k)$ forms a primitive Pythagorean triple, and so either
$$a^k=m^2-n^2,\qquad b=2mn,\qquad c^k=m^2+n^2,$$
or
$$a^k=2mn,\qquad b=m^2-n^2,\qquad c^k=m^2+n^2,$$
for some coprime positive integers $m$ and $n$ with $mn$ even. In the first case we see that
$$a^k+c^k=(m^2-n^2)+(m^2+n^2)=2m^2.$$
But by the main theorem of this article such a solution does not exist for $k\geq3$. In the second case we see that
$$a^k+c^k=(2mn)+(m^2+n^2)=(m+n)^2.$$
Again by the main theorem of that article such a solution does not exist for $k\geq4$. So either $k=2$ or $k=3$.
For $k=2$ it is a classical result of Fermat that the Diophantine equation
$$x^4-y^4=z^2,$$
has no nontrivial integral solutions. See also this question on math.stackexchange:
Solving $x^4-y^4=z^2$ .
This leaves only the case $k=3$, for which I do not see a slick solution.
Proposition: The Diophantine equation $x^6-y^6=z^2$ has no primitive integral solutions with $y\neq0$ even.
Proof. Let $x$, $y$ and $z$ be pairwise coprime integers such that $x^6-y^6=z^2$, with $y$ even and $|y|>0$ minimal. Then $(u,v,w)=(x^2,-y^2,z)$ satisfies
$$u^3+v^3=w^2,$$
and it is a classical result by Euler that any primitive solution is of one of the following three forms:
\begin{align}
(1)&\begin{cases}
u&=s(s+2t)(s^2-2st+4t^2)\\
v&=-4t(s-t)(s^2+st+t^2)\\
w&=\pm(s^2-2st-2t^2)(s^4+2s^3t+6s^2t^2-4st^3 + 4t^4)
\end{cases}
& 2\nmid s,\ 3\nmid(s-t),\\
\\
(2)&\begin{cases}
u&=s^4-4s^3t-6s^2t^2-4st^3 + t^4\\
v&=2(s^4+2s^3t+2st^3+t^4)\\
w&=3(s-t)(s+t)(s^4+2s^3t+6s^2t^2+2st^3+t^4)
\end{cases}
&2\nmid(s-t),\ 3\nmid(s-t),\\
\\
(3)&\begin{cases}
u&=-3s^4 + 6s^2t^2 + t^4\\
v&=3s^4+6s^2t^2-t^4\\
w&=6st(3s^4+t^4)
\end{cases}
&2\nmid(s-t),\ 3\nmid t,
\end{align}
after swapping $u$ and $v$ if necessary. In all three cases $s$ and $t$ are coprime integers. For a proof, see $\S 14.3.1$ of Henri's Cohen's Number Theory, Vol II,
Analytic and Modern Tools.
Because $w=z$ is odd the solution $(u,v,w)=(x^2,-y^2,z)$ cannot be of the form $(3)$. It cannot be of the form $(2)$ because $-y^2\equiv0\pmod{4}$, whereas $u\equiv1\pmod{4}$ and $v\equiv2\pmod{4}$. It folows that it is of the form $(1)$, and so
\begin{eqnarray}
x^2&=&s(s+2t)(s^2-2st+4t^4)&=&s(s^3+8t^3),\\
y^2&=&4t(s-t)(s^2+st+t^2)&=&4t(s^3-t^3)
\end{eqnarray}
for two coprime integers $s$ and $t$ with $s$ odd and $3\nmid(s-t)$. The factors in each product are pairwise coprime, and hence they are all perfect squares, say
$$s=S^2,\qquad t=T^2,\qquad s+2t=q^2,\qquad s^3-t^3=p^2,$$
for some integers $S$, $T$ , $p$ and $q$, not necessarily prime. Reducing mod $4$ shows that $t$ is even, and so we find
$$S^6-T^6=p^2,$$
with $S$, $T$ and $p$ pairwise coprime and $T$ even, where clearly $|S|<|x|$ and $|T|<|y|$. This contradicts the minimality of $|y|$, and so we conclude that there does not exist any primitive solution with $y\neq0$ even.$\qquad\square$
Together this finally shows that equation $(0)$ does not have any solutions in positive integers $a$, $b$ and $c$ for any integer $k>1$.
Best Answer
For anybody like me who is unable to view the solution at Diophantus Era begins! that Math Lover's question comment linked to because they're not a member of Brilliant and doesn't want to join, here is a solution.
First, apart from $x = 1$, which leads to the solution of $(1, 1, 3)$ you've already found, then both of the factors on the left are greater than $1$ and, thus, must be positive powers of $p$. This gives
$$x^2 + x + 1 \equiv 0 \pmod{p} \tag{1}\label{eq1A}$$
$$x^3 - x + 1 \equiv 0 \pmod{p} \tag{2}\label{eq2A}$$
Next, \eqref{eq2A} minus \eqref{eq1A} gives
$$x^3 - x^2 - 2x \equiv 0 \pmod{p} \implies x(x + 1)(x - 2) \equiv 0 \pmod{p} \tag{3}\label{eq3A}$$
This means $x \equiv 0 \pmod{p}$, $x \equiv -1 \pmod{p}$ or $x \equiv 2 \pmod{p}$. The first $2$ cases give in \eqref{eq1A} that $1 \equiv 0 \pmod{p} \implies p = 1$, which is not allowed. With the third case, \eqref{eq1A} gives $7 \equiv 0 \pmod{p} \implies p = 7$ is the only possibility.
Checking $x = 2$ itself shows it works to give on the left side $49 = 7^2$, so $(2, 2, 7)$ is another solution. Next, consider $x \gt 2$, so
$$x = 7z + 2 \tag{4}\label{eq4A}$$
for some integer $z \ge 1$. Substituting this into the $x^2 + x + 1$ factor, and letting it be equal to $7^m$ for some integer $m \ge 1$, gives
$$\begin{equation}\begin{aligned} 7^m & = (7z + 2)^2 + (7z + 2) + 1 \\ & = 49z^2 + 28z + 4 + 7z + 3 \\ & = 49z^2 + 35z + 7 \\ & = 7(7z^2 + 5z + 1) \end{aligned}\end{equation}\tag{5}\label{eq5A}$$
Thus, since $z \ge 1 \implies 7z^2 + 5z + 1 \gt 1$, then $7 \mid 7z^2 + 5z + 1$, so
$$5z + 1 \equiv 0 \pmod{7} \tag{6}\label{eq6A}$$
Next, letting $x^3 - x + 1 = 7^n$ for some integer $n \ge 1$ and using \eqref{eq4A} gives
$$\begin{equation}\begin{aligned} 7^n & = (7z + 2)^3 - (7z + 2) + 1 \\ & = (7)^3z^3 + 3(7^2)(2)z^2 + 3(7)(4)z + 8 - 7z - 1 \\ & = (7)^3z^3 + 6(7^2)z^2 + (11)(7)z + 7 \\ & = 7((7)^2z^3 + 6(7)z^2 + (11)z + 1) \end{aligned}\end{equation}\tag{7}\label{eq7A}$$
Similar to before, this results in
$$11z + 1 \equiv 0 \pmod{7} \implies 4z + 1 \equiv 0 \pmod{7} \tag{8}\label{eq8A}$$
Next, \eqref{eq6A} minus \eqref{eq8A} gives
$$z \equiv 0 \pmod{7} \tag{8}\label{eq9A}$$
However, this contradicts both \eqref{eq6A} and \eqref{eq8A} since it results in $1 \equiv 0 \pmod{7}$, showing there is no such positive integer $z$.
In summary, the only $2$ triplets $(x, y, p)$ that satisfy the equation are $(1, 1, 3)$ and $(2, 2, 7)$.