Inspired by Problem 4, IMO 2006

At first sight, there is no bound for variables. If you fix $p$ , for example, then how large $a$ might be, you can pick $b$ such that RHS balances LHS of the equation. Let us find a relation between $b$ and $p$ . This is reasonable because they are both bases. Then we might be able to bound one of variables.

Observe that $\gcd(b, p)=1$ . Look mod $p$ , we get $b^p \equiv 1 \pmod{p}$ and $b^{p-1}b \equiv b \equiv 1 \pmod{p}$ , using FLT . Thus, $p|b-1$ and we can apply LTE lemma in the following form.

Rearrange the equation to get $p^a(p^{2a+1}+1)=b^p-1.$

Note that the highest power of $p$ dividing LHS is $a$ , hence $a=v_p(b^p-1)=v_p(b-1)+v_p(p)=v_p(b-1)+1$ So $v_p(b-1)=a-1$ and there is an even integer $c$ , relatively prime to $p$ , such that $b=p^{a-1}c+1$ . Now we have the following equation $p^{3a+1}+p^a+1=(p^{a-1}c+1)^p \qquad (1)$

Now, from this equation, you see that the RHS would be greater than the LHS for almost any value of $a$ and $p$ and we have found our way to bound variables. We are going to narrow it down fairly quickly, i.e. prove that for most values RHS > LHS, and see what's left?

If $a=1$ , then $b \ge 3$ and original equation becomes $p^4+p+1=b^p \qquad(2)$ which results in $p^4+p+1=b^p \ge 3^p$ and $p<8$ , but none of $p=3, 5, 7$ give an integer for $b$ from equation $(2)$ .

Therefore, $a>1$ . According to the binomial theorem, RHS of the equation $(1)$ consists of a lot of terms with powers of $p$ greater than or equal to $a$ . Therefore, there mustn't be a power of $p$ in the RHS greater than or equal to the highest power of $p$ in the RHS, which is $3a+1$ . But note that for $p \ge 5$ and $a>1$ , since $c\ge2$ and $c^p>p^2$ the first term of binomial expansion becomes $p^{(a-1)p}c^p>p^{(a-1)p}p^2=p^{(a-1)p+2}$ and $(a-1)p+2\ge 5a-3 \ge 3a+1$ . This is a contradiction and leads to $p<5$ and the only remaining value $p=3$ . Thus, equation $(1)$ becomes $3^{3a+1}+3^a+1=(3^{a-1}c+1)^3=3^{3a-3}c^3+3^a c(3^{a-1}c+1)+1$ or $3^{2a-3} (81-c^3)=c(3^{a-1}c+1)-1$ LHS is positive. It follows that $81-c^3>0$ and $c\le 4$ . Note that $c$ is even and, thus, $c=2, 4$ . $c=2$ gives $77\times 3^{2a-3}=4\times 3^{a-1}+1,$ which is impossible, because LHS is always greater than RHS for $a>1$ . For $c=4$ we get $17\times 3^{2a-3}=16\times 3^{a-1}+3.$ For $a>2$ LHS of this equation becomes greater than its RHS and $a=2$ satisfies the equation. Therefore, the only solution to the given equation is $(a, b, p)=(2, 13, 3)$ .

$Let ~~f(p,a,b)=p^{3a+1} +p^a+1-b^p=0.\\ Try~ with~ smallest~ p=3.\\ Then~f(3,a,b)=3^ {3a+1}+3^a+1=b^3.\\ For~f(3,1,b)=85~~but~ this~ is~ not ~a ~cube~ of~ any~integer.\\ For~f(3,2,b)=2197, \sqrt[3]{2197}=~13,~~an~ integer.\\ 3+2+13=\huge~~\color{#D61F06}{18}.$

This is how I got the result. But it lacks the proof that this is the only solution.
Interesting observation.
$(5^{25}+5^8+1)-3125^5=0$ if we use accuracy up to 10 digits. Obviously it is not true since the term in the (.... ) has unit digit as 6 while the negative term has 5.
But when accuracy is 14 , we get that $(5^{25}+5^8+1)-3125^5=391000.$ And that changes in the last four digits as we go in for more accuracy.