Prove that there doesn't exist positive integers $a$ and $b$ such that $a^2+b$ and $b^2+a$ are both powers of $5$.
Assume on the contrary. Let $$a^2+b=5^x, \quad a+b^2=5^y$$ With $x,y\ge 1$. WLOG assume $x\ge y$ which means $a\ge b$ and $$a+b^2\mid a^2+b\implies a^2+b\equiv 0\pmod{a+b^2}$$
Since $a\equiv -b^2\pmod {a+b^2}$ we get $$a^2+b\equiv b^4+b\equiv 0\pmod{a+b^2}$$
Claim $(a,5)=(b,5)=1$. Furthermore $(a,b)=1$.
Proof Assume $5\mid a$ then immediately $5\mid b$. Let $a=5a_1,$ $ b=5b_1$ plugging this to the original equation $$5a_1^2+b_1=5^{x-1}$$ Now if $x=1$ we get a contradiction. So assume $x\ge 2$ clearly $5\mid b_1$ meaning $b_1=5b_2$. $$a_1^2+b_2=5^{x-2}$$ Again $x=2$ is impossible. So $a_1=5a_2...$ Cleary this can't go on forever hence $a_1=a_2=0$ a contradiction.
If $p\mid (a,b)$ then $p\mid 5$ meaning $p=5$ contradiction.
Back to our result $a+b^2\mid b(b^3+1)$ since $(a+b^2,b)=1$ we get $$a+b^2\mid b^3+1$$
We're pretty much done. we just need the following claim.
Claim $5\mid b+1$.
Proof We know $5\mid b^3+1$ Thus $$b^3\equiv -1\pmod 5\implies b^6\equiv 1\pmod 5$$ $\operatorname{ord}_5b\mid 6$ but $\operatorname{ord}_5b \mid 4$ so $\operatorname{ord}_5b=2$ or $1$. Surely it's not $1$. $$b^2\equiv 1\pmod 5 \implies b\equiv -1 \pmod 5$$ So $5\mid b+1$.
All our work till now was in order to use the LTE lemma (Lifting the exponent). Since $5\nmid b$ and $5\mid b+1$ we get $$\nu_5(a+b^2)\le \nu_5(b^3+1)=\nu_5(b+1)+\nu_5(3)=\nu_5(b+1)$$ In fact since the LHS is a power of $5$ we get $a+b^2\mid b+1 \implies b+b^2\le a+b^2\le b+1$. Hence $b\le 0$ A contradiction.
I just need a verification and if you see a part where I could've used a better method to do it -especially the first claim- please tell me.