The only solution of the Diophantine equation $(1) \;\; p^{q+1} + q^{p+1} = r^2$, where $p$ and $q$ are primes, is $=p=q=2$. Proof: We may (since $p$ and $q$ are symmetric in (1)) WLOG assume $p \geq q$. If both $p$ and $q$ are odd, then $2|p+1$ and $2|q+1$, yielding $r^2 = (p^{\frac{q+1}{2}})^2 + (q^{\frac{p+1}{2}})^2 \equiv 1 + 1 = 2 \pmod{4}$, a contradiction which implies $q=2$. If $p=2$, then $r=\pm 4$ by (1). Thus $(p,q,r)=(2,2,\pm4)$ is a solution of (1). Next assume $p>2$, i.e. $p$ is odd. Then according to (1) $(2) \;\; p^3 + 2^{p+1} = r^2$. The fact that $p+1$ is even means (2) can be expressed as $(r - 2^{\frac{p+1}{2}})(r + 2^{\frac{p+1}{2}}) = p^3$. Therefore, since $p$ is a prime, we must have ${\textstyle (r - 2^{\frac{p+1}{2}}, r + 2^{\frac{p+1}{2}}) = (k,\frac{p^3}{k})}$, where $k \in \{1,p\}$. Hence ${\textstyle (r + 2^{\frac{p+1}{2}}) - (r - 2^{\frac{p+1}{2}}) = \frac{p^3}{k} - k}$, i.e. ${\textstyle 2^{\frac{p+3}{2}} = \frac{p^3}{k} - k}$. If $k=p$, then ${\textstyle 2^{\frac{p+3}{2}} = p^2 - p}$, implying $p|2$, which is impossible since $p>2$. Consequently $k=1$, which means ${\textstyle 2^{\frac{p+3}{2}} = p^3 - 1 = (p-1)(p^2+p+1)}$, yielding $p-1=2^i$ and ${\textstyle p^2 + p + 1 = 2^{\frac{p+3}{2}-i}}$ for an ${\textstyle 0 < i < \frac{p+3}{2}}$. The fact that $p-1 < p^2 + p + 1$ means ${\textstyle 2^{\frac{p+3}{2}-2i} = \frac{p^2+p+1}{p-1} = p + 2 + \frac{3}{p-1}}$, implying $p-1|3$. Hence $p=4$, which is impossible since $p$ is an odd prime. Conclusion: The only solution of (1) is $p=q=2$.

By ($mod$ 4) we get that one of them is 2. In case of two even, it works, and that $p=q=2$ $a=4$. So we have case when one is odd. By ($mod$ 3) get that $a^2$ is type of $3k$ and that odd prime is type of $3l+2$. By ($mod$ 6) get that it’s impossible.

Pretty straightforward Diophantine here. We claim that the only triple $(p,q,x)$ of solutions is $(2,2,4)$. It is not hard to see that this triple of solutions indeed satisfies the given equation. Now we shall show that it is the only one. If there exists a solution besides the above one, then one of $p$ and $q$ must be odd. Say $p$ is odd. Then, \begin{align*} p^{q+1} + q^{p+1} &= x^2\\ p^{q+1} &= x^2 - q^{p+1}\\ p^{q+1} &= (x- q^{\frac{p+1}{2}})(x + q^{\frac{p+1}{2}}) \end{align*}Now, it is not hard to see that if $p\mid x- q^{\frac{p+1}{2}}$ and $p\mid x + q^{\frac{p+1}{2}}$, then $p\mid 2 \cdot q^{\frac{p+1}{2}}$. Since $p\neq 2$, this implies that we must have $p=q$. But then, \[p^{q+1}+q^{p+1}=2\cdot p^{p+1}\]which is clearly non-square since $\nu_2(p^{q+1}+q^{p+1})=1$. Thus, this is impossible and the smaller factor must be 1. Then, $x- q^{\frac{p+1}{2}}=1$ so, \begin{align*} 2\cdot q^{\frac{p+1}{2}}+1&=x + q^{\frac{p+1}{2}} = p^{q+1} \\ p^{q+1}-1 &= 2\cdot q^{\frac{p+1}{2}} \end{align*}But, by the Lifting the Exponent Lemma, \[\frac{p+1}{2} =\nu_{q}(p^{q+1}-1)=\nu_q(p-1)\]Thus, $p-1 \geq q^{\frac{p+1}{2}}$. But this doesn't work out since, \[p^{q+1}-1 \geq p^3 - 1> 2(p-1) \geq 2 \cdot q^{\frac{p+1}{2}}\]which is a clear contradiction. Thus, it is impossible for such a working pair of primes $(p,q)$ to exist, which finishes the problem.

quite straight forward