Such a sequence exists for $n=1,2,3,4$ and no other $n$. Since the existence of such a sequence for some $n$ implies the existence of such a sequence for all smaller $n$, it suffices to prove that $n=5$ is not possible and $n=4$ is possible.

Assume first that for $n=5$ there exists a sequence of positive integers $a_{1}, a_{2}, \ldots, a_{5}$ satisfying the conditions
$$\begin{aligned} & a_{2}^{2}+1=\left(a_{1}+1\right)\left(a_{3}+1\right), \\ & a_{3}^{2}+1=\left(a_{2}+1\right)\left(a_{4}+1\right), \\ & a_{4}^{2}+1=\left(a_{3}+1\right)\left(a_{5}+1\right) . \end{aligned}$$
Assume $a_{1}$ is odd, then $a_{2}$ has to be odd as well and as then $a_{2}^{2}+1 \equiv 2 \bmod 4, a_{3}$ has to be even. But this is a contradiction, since then the even number $a_{2}+1$ cannot divide the odd number $a_{3}^{2}+1$.

Hence $a_{1}$ is even.
If $a_{2}$ is odd, $a_{3}^{2}+1$ is even (as a multiple of $a_{2}+1$ ) and hence $a_{3}$ is odd, too. Similarly we must have $a_{4}$ odd as well. But then $a_{3}^{2}+1$ is a product of two even numbers $\left(a_{2}+1\right)\left(a_{4}+1\right)$ and thus is divisible by 4 , which is a contradiction as for odd $a_{3}$ we have $a_{3}^{2}+1 \equiv 2 \bmod 4$.

Hence $a_{2}$ is even. Furthermore $a_{3}+1$ divides the odd number $a_{2}^{2}+1$ and so $a_{3}$ is even. Similarly, $a_{4}$ and $a_{5}$ are even as well.

Now set $x=a_{2}$ and $y=a_{3}$. From the given condition we get $(x+1) \mid\left(y^{2}+1\right)$ and $(y+1) \mid\left(x^{2}+1\right)$. We will prove that there is no pair of positive even numbers $(x, y)$ satisfying these two conditions, thus yielding a contradiction to the assumption.

Assume there exists a pair ( $x_{0}, y_{0}$ ) of positive even numbers satisfying the two conditions $\left(x_{0}+1\right) \mid\left(y_{0}^{2}+1\right)$ and $\left(y_{0}+1\right) \mid\left(x_{0}^{2}+1\right)$.
Then one has $\left(x_{0}+1\right) \mid\left(y_{0}^{2}+1+x_{0}^{2}-1\right)$, i.e., $\left(x_{0}+1\right) \mid\left(x_{0}^{2}+y_{0}^{2}\right)$, and similarly $\left(y_{0}+1\right) \mid\left(x_{0}^{2}+y_{0}^{2}\right)$. Any common divisor $d$ of $x_{0}+1$ and $y_{0}+1$ must hence also divide the number $\left(x_{0}^{2}+1\right)+\left(y_{0}^{2}+1\right)-\left(x_{0}^{2}+y_{0}^{2}\right)=2$. But as $x_{0}+1$ and $y_{0}+1$ are both odd, we must have $d=1$. Thus $x_{0}+1$ and $y_{0}+1$ are relatively prime and therefore there exists a positive integer $k$ such that
$$k(x+1)(y+1)=x^{2}+y^{2}$$
has the solution $\left(x_{0}, y_{0}\right)$. We will show that the latter equation has no solution $(x, y)$ in positive even numbers.

Assume there is a solution. Pick the solution $\left(x_{1}, y_{1}\right)$ with the smallest sum $x_{1}+y_{1}$ and assume $x_{1} \geq y_{1}$. Then $x_{1}$ is a solution to the quadratic equation
$$x^{2}-k\left(y_{1}+1\right) x+y_{1}^{2}-k\left(y_{1}+1\right)=0$$
Let $x_{2}$ be the second solution, which by Vieta's theorem fulfills $x_{1}+x_{2}=k\left(y_{1}+1\right)$ and $x_{1} x_{2}=y_{1}^{2}-k\left(y_{1}+1\right)$. If $x_{2}=0$, the second equation implies $y_{1}^{2}=k\left(y_{1}+1\right)$, which is impossible, as $y_{1}+1>1$ cannot divide the relatively prime number $y_{1}^{2}$. Therefore $x_{2} \neq 0$.

Also we get $\left(x_{1}+1\right)\left(x_{2}+1\right)=x_{1} x_{2}+x_{1}+x_{2}+1=y_{1}^{2}+1$ which is odd, and hence $x_{2}$ must be even and positive. Also we have $x_{2}+1=\frac{y_{1}^{2}+1}{x_{1}+1} \leq \frac{y_{1}^{2}+1}{y_{1}+1} \leq y_{1} \leq x_{1}$. But this means that the pair ( $x^{\prime}, y^{\prime}$ ) with $x^{\prime}=y_{1}$ and $y^{\prime}=x_{2}$ is another solution of $k(x+1)(y+1)=x^{2}+y^{2}$ in even positive numbers with $x^{\prime}+y^{\prime}<x_{1}+y_{1}$, a contradiction.

Therefore we must have $n \leq 4$.
When $n=4$, a possible example of a sequence is $a_{1}=4, a_{2}=33, a_{3}=217$ and $a_{4}=1384$.

It is easy to check that for $n=4$ the sequence $a_{1}=4, a_{2}=33, a_{3}=217$ and $a_{4}=1384$ is possible.

Now assume there is a sequence with $n \geq 5$. Then we have in particular
$$\begin{aligned} & a_{2}^{2}+1=\left(a_{1}+1\right)\left(a_{3}+1\right), \\ & a_{3}^{2}+1=\left(a_{2}+1\right)\left(a_{4}+1\right), \\ & a_{4}^{2}+1=\left(a_{3}+1\right)\left(a_{5}+1\right) . \end{aligned}$$
Also assume without loss of generality that among all such quintuples $\left(a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\right)$ we have chosen one with minimal $a_{1}$.

One shows quickly the following fact:
$$\begin{align*} & \text{ If three positive integers } x, y, z \text{ fulfill } y^{2}+1=(x+1)(z+1) \text{ and if } y \text{ is even, then } \\ & x \text{ and } z \text{ are even as well and either } x<y<z \text{ or } z<y<x \text{ holds. } \tag{1} \end{align*}$$
Indeed, the first part is obvious and from $x<y$ we conclude
$$z+1=\frac{y^{2}+1}{x+1} \geq \frac{y^{2}+1}{y}>y$$
and similarly in the other case.

Now, if $a_{3}$ was odd, then $\left(a_{2}+1\right)\left(a_{4}+1\right)=a_{3}^{2}+1 \equiv 2 \bmod 4$ would imply that one of $a_{2}$ or $a_{4}$ is even, this contradicts (1). Thus $a_{3}$ and hence also $a_{1}, a_{2}, a_{4}$ and $a_{5}$ are even. According to (1), one has $a_{1}<a_{2}<a_{3}<a_{4}<a_{5}$ or $a_{1}>a_{2}>a_{3}>a_{4}>a_{5}$ but due to the minimality of $a_{1}$ the first series of inequalities must hold.

Consider the identity
$$\left(a_{3}+1\right)\left(a_{1}+a_{3}\right)=a_{3}^{2}-1+\left(a_{1}+1\right)\left(a_{3}+1\right)=a_{2}^{2}+a_{3}^{2}=a_{2}^{2}-1+\left(a_{2}+1\right)\left(a_{4}+1\right)=\left(a_{2}+1\right)\left(a_{2}+a_{4}\right) .$$
Any common divisor of the two odd numbers $a_{2}+1$ and $a_{3}+1$ must also divide ( $a_{2}+1$ )( $a_{4}+ 1)-\left(a_{3}+1\right)\left(a_{3}-1\right)=2$, so these numbers are relatively prime. Hence the last identity shows that $a_{1}+a_{3}$ must be a multiple of $a_{2}+1$, i.e. there is an integer $k$ such that
$$\begin{equation*} a_{1}+a_{3}=k\left(a_{2}+1\right) . \tag{2} \end{equation*}$$
Now set $a_{0}=k\left(a_{1}+1\right)-a_{2}$. This is an integer and we have
$$\begin{aligned} \left(a_{0}+1\right)\left(a_{2}+1\right) & =k\left(a_{1}+1\right)\left(a_{2}+1\right)-\left(a_{2}-1\right)\left(a_{2}+1\right) \\ & =\left(a_{1}+1\right)\left(a_{1}+a_{3}\right)-\left(a_{1}+1\right)\left(a_{3}+1\right)+2 \\ & =\left(a_{1}+1\right)\left(a_{1}-1\right)+2=a_{1}^{2}+1 \end{aligned}$$
Thus $a_{0} \geq 0$. If $a_{0}>0$, then by (1) we would have $a_{0}<a_{1}<a_{2}$ and then the quintuple ( $a_{0}, a_{1}, a_{2}, a_{3}, a_{4}$ ) would contradict the minimality of $a_{1}$.

Hence $a_{0}=0$, implying $a_{2}=a_{1}^{2}$. But also $a_{2}=k\left(a_{1}+1\right)$, which finally contradicts the fact that $a_{1}+1>1$ is relatively prime to $a_{1}^{2}$ and thus cannot be a divisior of this number.

Hence $n \geq 5$ is not possible.