Positive integer solutions to $ p+q^{n}=x^2$

Question

Consider two prime numbers $p$ and $q$ such that $$ p+q^2=r^2, $$ and $r\in\mathbb N$. It is not difficult to figure out that for any $n\in\mathbb N$ and $x\in\mathbb N$ there are no solutions of $$ p^{2}+q^{n}=x^2. $$ What about $$ p+q^{n}=x^2, $$ with $n>2$? So far I was only able to prove that there are no positive integer solutions if $q=2$. The question is a variation of a problem in the Italian 2019 mathematical Olympiads.

Answer 1

$p + q^2 = r^2$ means $p = r^2 - q^2 = (r+q)(r-q)$. Since $p$ is prime, this implies $r-q=1$ and $r+q=p$, i.e. $p=2q+1$.

If $n > 2$ is even, $p + q^n = r^2$ means $p = (r+q^{n/2})(r-q^{n/2})$, so $r=q^{n/2}+1$ and $p = r + q^{n/2} = 2 q^{n/2}+1$.

If $n$ is divisible by $4$, there can't be solutions with $q \ne 3$, as $2 q^{n/2}+1$ would be divisible by $3$. For $n$ even but not divisible by $4$, I would expect solutions to exist.

Some examples:

$$\eqalign{5 + 2^2 &= 3^2\cr 19 + 3^4 &= 10^2\cr 17 + 2^6 &= 9^2\cr 163 + 3^8 &= 82^2\cr 487 + 3^{10} &= 244^2\cr 1459 + 3^{12} &= 730^2\cr 257 + 2^{14} &= 129^2\cr 39367 + 3^{18} &= 19684^2\cr 5264950288664609755476607823 + 311^{22} &= 2632475144332304877738303912^2\cr 2441406251 + 5^{26} &= 1220703126^2\cr 65537 + 2^{30} &= 32769^2\cr }$$

Answer 2

Regarding your last request of

$$p + q^n = x^2 \tag{1}\label{eq1}$$

having any solutions for $n \gt 2$, and not including $q = 2$. An example is $p = 19$, $q = 3$, $n = 4$ and $x = 10$ since $19 + 3^4 = 19 + 81 = 100 = 10^2$.