*Answer:* $(2^0 + 1, 2^1 + 1, 2^2 + 1, 2^3 + 1, 2^4 + 1)$.

Clearly $(2^0 + 1, 2^1 + 1, 2^2 + 1, 2^3 + 1, 2^4 + 1)$ is a perfect tuple with length 5. Show in the rest that there are no other perfect tuples with length 5 or larger.

For that, let $(a^m + 1, a^{m+1} + 1, \dots, a^n + 1)$ be an arbitrary perfect tuple with length at least 5. There must exist at least two odd exponents among $m, m+1, \dots, n$; let $k$ and $k+2$ be the two largest odd exponents. As $a^k + 1$ and $a^{k+2} + 1$ are prime powers while having a common divisor $a+1$, these two integers must be powers of the same prime $q$. Thus the larger of them, $a^{k+2} + 1$, is divisible by the smaller one, $a^k + 1$, which shows that $a^k + 1$ divides also the difference $a^2 \cdot (a^k + 1) - (a^{k+2} + 1) = a^2 - 1$. Hence $a^k + 1 \le a^2 - 1$, implying $k < 2$. So $k = 1$ as $k$ is odd. By choice of $k$, the only odd exponents in our perfect tuple are 1 and 3 and the tuple is of the form $(a^0 + 1, a^1 + 1, a^2 + 1, a^3 + 1, a^4 + 1)$.

As $a+1$ and $a^3+1$ are powers of the same prime number $q$, also the ratio $\frac{a^3+1}{a+1} = a^2-a+1$ is a power of $q$. Note that $a^2-a+1 \ge 2a-a+1 = a+1$ by $a \ge 2$, hence $a^2-a+1$ is divisible by $a+1$. Thus the difference $(a^2-a+1)-(a+1)(a-2) = 3$ is divisible by $a+1$. This filters out the only possibility $a=2$.