According to the given information $a \neq b$ are two natural numbers and there is a prime $p$ and a natural number $k$ satisfying $(1) \;\; a^2 + b \mid b^2 + a$ and $(2) \;\; b^2 + a = p^k$. Combining conditions (1) and (2) we obtain $a^2 + b \mid p^k$, implying there is a natural number $m \leq k$ s.t. $(3) \;\; a^2 + b = p^m$. Subtracting equation (3) from equation (2), the result is $(4) \;\; (b - a)(b + a - 1) = p^m(p^{k-m} - 1)$. Hence $a<b$ and $m<k$ by equation (4). Assume $p^m \mid b-a$ or $p^m \mid b + a - 1$. Consequently, since $b+a-1>b-a$, we have $p^m \leq b + a - 1$, which according to equation (3) yields $b + a - 1 \geq a^2 + b$, i.e. $a^2 - a + 1 \leq 0$, which is impossible since $a \geq 1$. This contradiction combined with equation (4) give us $(5) \;\; p \mid b - a$, $(6) \;\; p \mid b + a - 1$. Subtracting (5) from (6) and adding (5) and (6) we obtain $p \mid 2a - 1$ and $p \mid 2b - 1$ respectively. Hence $p$ is odd, which according to equation (2) give us $0 \equiv 4p^k = (2b)^2 + 2 \cdot (2a) \equiv 1^2 + 2 \cdot 1 = 1 + 2 = 3 \pmod{p}$, i.e. $3|p$. Therefore $p=3$, which inserted in equation (2)-(4) give us $(7) \;\; b^2 + a = 3^k$, $(8) \;\; a^2 + b = 3^m$, $(9) \;\; (b - a)(b + a - 1) = 3^m(3^{k-m} - 1)$. Observe that $3^k = b^2 + a \geq 2^2 + 1 = 4+1 = 5 > 3$, yielding $k \geq 2$. Next assume $9 \mid b-a$ and $9 \mid b+a-1$. Hence $9 \mid (b + a - 1) \pm (b - a)$, which means $9 \mid 2a - 1$ and $9 \mid 2b - 1$. Thus we have $a \equiv b \equiv 5 \pmod{9}$, which according to equation (7) (since $k \geq 2$) yields $0 \equiv 3^k = b^2 + a \equiv 5^2 + 5 = 25 + 5 = 30 \equiv 3 \pmod{9}$, a contradiction implying $9 \nmid b-a$ or $9 \nmid b+a-1$. Hence by equation (9) we obtain $3^{m-1} \mid b-a$ or $3^{m-1} \mid b+a-1$. Hence there are two natural numbers $s$ and $t$ s.t. $(10) \;\; b - a = 3^{m-1}s$ or $(11) \;\; b+a-1 = 3^{m-1}t$, where $s,t \in \{1,2\}$ (we know that $b-a < b+a-1 < p^m = 3^m$). Thus we have the following two cases: Case 1: $b-a=3^{m-1}s$. If $s=2$, then $b-a$ is even, yielding $a^2 + b$ is even, contradicting equation (8). Consequently $s=1$, which give us $b-a = 3^{m-1}$, which combined with equation (8) yields $3(b - a) = 3^m = a^2 + b$. This means $2b = a^2 + 3a$, which according to equation (8) result in $2 \cdot 3^m = 2a^2 + 2b = 2a^2 + (a^2 + 3a) = 3a(a+1)$, i.e. $(12) \;\; a(a+1) = 2 \cdot 3^{m-1}$. Therefore, since $GCD(a,a+1)=1$, we have $3^{m-1} \mid a$ or $3^{m-1} \mid a+1$, which combined with equation (12) yields $a \leq 2$. Thus we obtain $3^{m-1}=1$ when $a=1$ and $3^{m-1}=3$ when $a=2$. In other words, since $2b=a^2+3a$, we obtain $(a,b,m)=(1,2,1)$ or $(a,b,m)=(2,5,2)$, yielding $b^2+a=2^2+1=4+1=5$ and $b^2+a=5^2+2=25+2=27=3^3$. Hence by equation (7) the only solution of equations (7)-(8) is $(a,b,m,k)=(2,5,2,3)$. Case 2: $b+a-1=3^{m-1}t$. Then according to equation (8) $(a^2 + b)t = 3^m \cdot t = 3(b + a - 1)$, i.e. $(13) \;\; (3 - t)b = ta^2 - 3a + 3$. If $t=1$, then $a^2 - 3a + 3$ is even by equation (13), which is impossible since $a^2 - 3a$ is even. Consequently $t=2$ by contradiction, which inserted in equation (13) give us $(14) \;\; b = 2a^2 - 3a + 3$. Combining equations (8) and (14) we find that $3^m = a^2 + b = a^2 + (2a^2 - 3a + 3) = 3(a^2 - a + 1)$, i.e. $(15) \;\; a^2 - a + 1 = 3^{m-1}$. The fact that $9 \nmid a^2 - a + 1$ combined with equation (15) yields $m \leq 2$. Thus $(a,b)=(1,2)$ or $(a,b)=(2,5)$ by equations (15) and (14). From case 1 we know that the only solution is $(a,b)=(2,5)$ with $(k,m)=(3,2)$. Conclusion: The only pair of distinct natural numbers $(a,b)$ satisfying conditions (1) and (2) is $(a,b)=(2,5)$.