Lemma. If $P$ is a polynomial with integer coefficients and a square leading coefficient which for infinitely many numbers $n$, $P(n)$ is a square number, then there exists a $Q \in \mathbb{Z}[x]$ which
$$P(x) = Q(x)^2$$
Assume $P(x)$ has the following representation
$$P(x) = c_n x^n + c_{n-1} x^{n-1} + \dots + c_0$$
According to the assumptions, there exists an integer number $c$ which
$$c^2 = c_n \implies P(x) = c^2 x^n + c_{n-1} x^{n-1} + \dots + c_0$$
Set $k = as^2 + 2s$.
$$ak + 1 = a(as^2 + 2s) + 1 = a^2s^2 + 2as + 1 = (as + 1)^2$$
Since $ak+1$ is a square number by the assumption one of the numbers $a_i k+1$ must be a square.
For every integer $1 \le i \le n$ define polynomial $P_i(s)$ by the following equation.
$$P_i(s) = a_i(as^2 + 2s) + 1 = aa_i s^2 + 2a_i s + 1 \quad (1)$$
So $P_i(s)$ is a square for an integer $i$. Now put $s, s+1, \dots, s+n$ in these polynomials. By applying pigeonhole principle two of these $i$'s are equal so $P_i(s+t), P_i(s+k)$ are square numbers for some $0 \le t, k \le n$. If $s$ goes to infinity, for each $s$ there is a triple $(i, j, k)$, which $P_i(s+t), P_i(s+k)$ are square numbers. Since there are only finite number of these triples so there are infinite numbers $s$ such that they have same triples.
Define polynomial $P$ such that
$$P(s) = P_i(s+t)P_i(s+k).$$
Since $P_i(s+t), P_i(s+k)$ are squares so $P(s)$ is a square number too. Also because $P_i(s+t), P_i(s+k)$ are polynomials of degree 2 with the same leading coefficient $aa_i$ then $P(s)$ is a polynomial of degree 4 with a square leading coefficient.
The lemma suggests that there exists a polynomial of degree 2 such that
$$P(x) = Q(x)^2$$
also
$$P(x) = P_i(x+t)P_i(x+k)$$
by putting together
$$P_i(x+t)P_i(x+k) = Q(x)^2.$$
Let $D(x)$ be the greatest common divisor of $P_i(x+t), P_i(x+k)$ so
$$P_i(x+t) = D(x)A(x)^2, \quad P_i(x+k) = D(x)B(x)^2$$
Notice that $P_i(x+t), P_i(x+k)$ are polynomials of degree 2 so either of $A, B$ are constant or of degree one. If they are constant $P_i(x+t), P_i(x+k)$ should be a multiple of each other but they have same leading coefficients then they are equal which is a contradiction because we chose different $t, k$. So $D$ should be a constant but for infinite integers $s$, $D(s)A(s)^2$ is a square number so for some $s$, which $A(s) \ne 0$ because $a, a_i$ are positive integers and then $P_i$ can't be zero.
$$\begin{align*} D &= d^2 \implies P_i(x+t) = (dA(x))^2, \quad P_i(x+k) = (dB(x))^2 \\ \implies (mx+p)^2 &= P_i(x) \stackrel{(1)}{=} aa_i x^2 + 2a_i x + 1 \end{align*}$$
so $p^2 = 1$. If $p = -1$ take
$$(-mx - p)^2 = (mx + p)^2$$
so without loss of generality, take $p = 1$
$$aa_i x^2 + 2a_i x + 1 = (mx + 1)^2$$
by comparing coefficients
$$aa_i = m^2, \ 2a_i = 2m$$
from the last equation $a_i = m$ so by putting this into the first equation
$$aa_i = a_i^2 \underset{a_i \neq 0}{\stackrel{a_i \neq 0}{\Rightarrow}} a = a_i$$
then
$$a \in \{a_1, a_2, \dots, a_n\}$$
and we are done.

As the first solution take $k = as^2 + 2s$. So for an integer $i$, $P_i(s)$ is a square number. i.e.
$$P_i(s) = a_i(as^2 + 2s) + 1 = aa_i s^2 + 2a_i s + 1$$
is a square.
$$(a_i s + 1)^2 = a_i^2 s^2 + 2a_i s + 1$$
by subtracting these two equalities
$$\begin{aligned} P_i(s) - (a_i s + 1)^2 &= (aa_i s^2 + 2a_i s + 1) - (a_i^2 s^2 + 2a_i s + 1) = a_i(a - a_i)s^2 \\ implies P_i(s) - (a_i s + 1)^2 &= a_i(a - a_i)s^2 \end{aligned} \qquad (2)$$
By the assumption there is a positive integer $n_i$ which
$$n_i^2 = P_i(s)$$
so by putting this in (2)
$$a_i(a - a_i)s^2 = P_i(s) - (a_i s + 1)^2 = n_i^2 - (a_i s + 1)^2 = (n_i - a_i s - 1)(n_i + a_i s + 1)$$
Let $s$ be a prime number. Assume that $a \neq a_i$. By the equality $s^2$ divides the right hand side.
$$s^2 \mid (n_i - a_i s - 1)(n_i + a_i s + 1)$$
If $s$ divides both of the numbers in the parentheses
$$s \mid n_i - a_i s - 1, \quad s \mid n_i + a_i s + 1$$
By subtracting
$$s \mid 2a_i s + 2 \implies s \mid 2$$
So $s$ only divides one of these numbers and by assuming $a - a_i \neq 0$
$$s^2 \leq \max(n_i - a_i s - 1, n_i + a_i s + 1) \underset{a_i \geq 0}{\stackrel{a_i \geq 0}{\Rightarrow}} s^2 \leq n_i + a_i s + 1$$
Notice that the right hand side is approximately a polynomial of degree one in terms of $s$ and it's a contradiction
$$\begin{aligned} s^2 &\le n_i + a_i s + 1 = \sqrt{aa_i s^2 + 2a_i s + 1} + a_i s + 1 \\ implies s^2 - a_i s - 1 &\le \sqrt{aa_i s^2 + 2a_i s + 1} \end{aligned}$$
For large enough prime number $s$ the left hand side is a positive integer so by taking square of both sides.
$$(s^2 - a_i s - 1)^2 \le a_i s^2 - 2a_i s - 1 \implies (s^2 - a_i s - 1)^2 - a_i s^2 - 2a_i s - 1 \le 0$$
This inequality says that a polynomial of degree four with a positive leading coefficient is a negative number for infinite prime numbers which is a contradiction. Notice that there are only $n$ possibilities for $i$ so for infinite prime numbers there is an integer $i$ that these equalities hold. ■