The answer is no.
Suppose that there exist a sequence $(a_n)$ of positive integers satisfying the given condition. We will show that this will lead to a contradiction.
For each $n \ge 2$ define $b_n = a_{n+1} - a_n$. Then, by assumption, for $n \ge 2$ we get $b_n = \sqrt{a_n + a_{n-1}}$ so that we have
$$b_{n+1}^2 - b_n^2 = (a_{n+1} + a_n) - (a_n + a_{n-1}) = (a_{n+1} - a_n) + (a_n - a_{n-1}) = b_n + b_{n-1}.$$
Since each $a_n$ is a positive integer we see that $b_n$ is positive integer for $n \ge 2$ and the sequence $(b_n)$ is strictly increasing for $n \ge 3$. Thus $b_n + b_{n-1} = (b_n - b_{n-1})(b_n + b_{n-1}) \ge (b_{n+1} + b_n)$, whence $b_{n-1} \ge b_{n+1}$ - a contradiction to increasing of the sequence $(b_i)$.
Thus we conclude that there exists no sequence $(a_n)$ of positive integers satisfying the given condition of the problem.

Suppose that such a sequence exists. We will calculate its members one by one and get a contradiction.
From the equality $a_3 = a_2 + \sqrt{a_2 + a_1}$ it follows that $a_3 > a_2$. Denote positive integers $\sqrt{a_3 + a_2}$ by $b$ and $a_3$ by $a$, then we have $\sqrt{2a} > b$. Since $a_4 = a + b$ and $a_5 = a + b + \sqrt{2a + b}$ are positive integers, then $\sqrt{2a + b}$ is positive integer.
Consider $a_6 = a+b+\sqrt{2a+b}+\sqrt{2a+2b+\sqrt{2a+b}}$. Number $c = \sqrt{2a+2b+\sqrt{2a+b}}$ must be positive integer, obviously it is greater than $\sqrt{2a+b}$. But
$$(\sqrt{2a+b}+1)^2 = 2a+b+2\sqrt{2a+b}+1 = 2a+2b+\sqrt{2a+b}+(\sqrt{2a+b}-b)+1 > c^2.$$
So $\sqrt{2a+b} < c < \sqrt{2a+b} + 1$ which is impossible.

We will show that there is no sequence $(a_n)$ of positive integers which consists of $N > 5$ members and satisfies
$$a_{n+2} = a_{n+1} + \sqrt{a_{n+1} + a_n} \quad (1)$$
for all $n = 1, 2, \dots, N-2$. Moreover, we will describe all such sequences with five members.
Since every $a_i$ is a positive integer it follows from (1) that there exists such positive integer $k$ (obviously $k$ depends on $n$) that
$$a_{n+1} + a_n = k^2 \quad (2)$$
From (1) we have $(a_{n+2} - a_{n+1})^2 = a_{n+1} + a_n$, consider this equality as a quadratic equation with respect to $a_{n+1}$,
$$a_{n+1}^2 - (2a_{n+2} + 1)a_{n+1} + a_{n+2}^2 - a_n = 0.$$
Obviously its solutions are $(a_{n+1})_{1/2} = \frac{2a_{n+2}+1\pm\sqrt{D}}{2}$, where
$$D = 4(a_n + a_{n+2}) + 1. \quad (3)$$
Since $a_{n+2} > a_{n+1}$ we have
$$a_{n+1} = \frac{2a_{n+2} + 1 - \sqrt{D}}{2}.$$
From the last equality, using that $a_{n+1}$ and $a_{n+2}$ are positive integers, we conclude that $D$ is a square of some odd number i.e. $D = (2m+1)^2$ for some positive integer $m \in \mathbb{N}$, substitute this into (3):
$$a_n + a_{n+2} = m(m+1). \quad (4)$$
Now adding $a_n$ to both sides of (1) and using (2) and (4) we get $m(m+1) = k^2 + k$ whence $m = k$. So
$$\begin{cases} a_n + a_{n+1} = k^2 \\ a_n + a_{n+2} = k^2 + k \end{cases} \quad (5)$$
for some positive integer $k$ (recall that $k$ depends on $n$).
Write equations (5) for $n=2$ and $n=3$, then for some positive integers $k$ and $l$ we get
$$\left\{ \begin{array}{l} a_2 + a_3 = k^2 \\ a_2 + a_4 = k^2 + k \\ a_3 + a_4 = l^2 \\ a_3 + a_5 = l^2 + l \end{array} \right. . \quad (6)$$
Solution of this linear system is
$$a_2 = \frac{2k^2 - l^2 + k}{2}, \quad a_3 = \frac{l^2 - k}{2}, \quad a_4 = \frac{l^2 + k}{2}, \quad a_5 = \frac{l^2 + 2l + k}{2}. \quad (7)$$
From $a_2 < a_4$ we obtain $k^2 < l^2$ hence $k < l$.
Consider $a_6$:
$$a_6 = a_5 + \sqrt{a_5 + a_4} = a_5 + \sqrt{l^2 + l + k}.$$
Since $0 < k < l$ we have $l^2 < l^2 + l + k < (l+1)^2$. So $a_6$ cannot be integer i.e. there is no such sequence with six or more members.
To find all required sequences with five members we must find positive integers $a_2, a_3, a_4$ and $a_5$ which satisfy (7) for some positive integers $k < l$. Its clear that $k$ and $l$ must be of the same parity. Vice versa, let positive integers $k,l$ be of the same parity and satisfy $k<l$ then from (7) we get integers $a_2, a_3, a_4$ and $a_5$ then $a_1=(a_3-a_2)^2-a_2$ and it remains to verify that $a_1$ and $a_2$ are positive i.e. $2k^2+k>l^2$ and $2(l^2-k^2-k)^2>2k^2-l^2+k$.

It is easy to see that $(a_n)$ is increasing for large enough $n$. Hence
$$a_{n+1} < a_n + \sqrt{2a_n} \qquad (1)$$
and
$$a_n < a_{n-1} + \sqrt{2a_{n-1}} \qquad (2)$$
Lets define $b_n = a_n + a_{n-1}$. Using AM-QM inequality we have
$$\frac{\sqrt{2a_n} + \sqrt{2a_{n-1}}}{2} \le \sqrt{\frac{2a_n + 2a_{n-1}}{2}} \qquad (3)$$
Adding (1), (2) and (3):
$$b_{n+1} < b_n + \sqrt{2a_n} + \sqrt{2a_{n-1}} \le b_n + 2\sqrt{b_n}.$$
Let $b_n = m^2$. Since $(b_n)$ is increasing for large enough $n$ we have:
$$m^2 < b_{n+1} < m^2 + 2m < (m+1)^2.$$
So, $b_{n+1}$ can't be a perfect square, so we get contradiction.