Proving that for any three integers $a,b,c$ there exists a positive integer $n$ such that $\sqrt{n^3+an^2+bn+c}$ is not an integer

I'm not sure offhand how to solve the problem using your particular approach. Instead, here is an alternate method. First, let

$$f(n) = n^3 + an^2 + bn + c \tag{1}\label{eq1A}$$

Note all perfect squares are congruent to either $0$ or $1$ modulo $4$. Thus, the difference of any $2$ perfect squares will be congruent to $-1$, $0$ or $1$ modulo $4$. In particular, it'll never be congruent to $2$ modulo $4$, i.e., have only one factor of $2$.

You don't mention the integers $a$, $b$ and $c$ need to be positive, so there may be some values of $n$ where $f(n)$ in \eqref{eq1A} is negative and, thus, its square root would not even be a real value. In any case, there will always be a positive integer $n_0$ such that for all $n \ge n_0$ we get $f(n) \ge 0$.

For any integers $n_1 \ge n_0$ and $d \gt 0$, we get

$$\begin{equation}\begin{aligned} f(n_1 + d) - f(n_1) & = ((n_1 + d)^3 + a(n_1 + d)^2 + b(n_1 + d) + c) \\ & \; \; \; \; - (n_1^3 + an_1^2 + bn_1 + c) \\ & = (n_1^3 + 3n_1^2d + 3n_1d^2 + d^3 + an_1^2 + 2an_1d + ad^2 \\ & \; \; \; \; \; \; + bn_1 + bd + c) - (n_1^3 + an_1^2 + bn_1 + c) \\ & = 3n_1^2d + 3n_1d^2 + d^3 + 2an_1d + ad^2 + bd \\ & = d(3n_1^2 + 3n_1d + d^2 + 2an_1 + ad + b) \end{aligned}\end{equation}\tag{2}\label{eq2A}$$

Now, consider $d$ to be any even positive integer with just one factor of $2$, e.g., $d = 2$. All of the terms inside the brackets in \eqref{eq2A} would then be even except for $3n_1^2 + b$. If $b$ is even, choose an odd $n_1$, else if $b$ is odd, choose an even $n_1$. This makes the part inside the brackets odd, so the right side of \eqref{eq2A} has just one factor of $2$, which means it's congruent to $2$ modulo $4$. Thus, at least one of $f(n_1)$ and $f(n_1 + d)$ cannot be a perfect square, so its square root would not be an integer.