I am not sure how elementary you want your proof to be, but here is a proof that uses elliptic curves...
Suppose that there are $x,y\in\mathbb{Z}$ such that $x>0$ and
$$y^2=x(x+1)(x+2)(x+3)(x+4)(x+5).$$
If we put $t=x+2+\frac{1}{2}$, then we have
$$y^2=(t-5/2)(t-3/2)(t-1/2)(t+1/2)(t+3/2)(t+5/2)=\left(t^2-\frac{1}{4}\right)\left(t^2-\frac{9}{4}\right)\left(t^2-\frac{25}{4}\right),$$
or, equivalently,
$$4^3y^2 = (4t^2-1)(4t^2-9)(4t^2-25).$$
If we put $U=2^3y$ and $V=4t^2$, then we have a solution for the equation
$$U^2=(V-1)(V-9)(V-25)=V^3 - 35V^2 + 259V - 225.$$
This defines an elliptic curve $E/\mathbb{Q}$, and we can use standard techniques to calculate the rank of the group of rational points $E(\mathbb{Q})$. This method ($2$-descent) shows that the rank of the curve is $0$, and one can easily separately show that the torsion subgroup is $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$. It follows that the only points on $E(\mathbb{Q})$ are the trivial points $(V,U)=(1,0)$, $(9,0)$ and $(25,0)$, plus the point ``at infinity'' on the curve. These correspond to $t$-values $t=\pm 1/2$, $\pm 3/2$ and $\pm 5/2$, and therefore do not give any integer values of $x$ with $x> 0$. Hence, there are no integer solutions to our original equation.
There is probably some elementary argument that shows that $U^2=(V-1)(V-9)(V-25)$ only has $3$ solutions, but I can't think of one right away.
Your technique should have worked, but if you don't know which expansions to do first you can get yourself in a tangle of algebra and make silly mistakes that bring the whole thing crashing down.
The way I reasoned was, well, I have four numbers multiplied together, and I want it to be two numbers of the same size multiplied together. So I'll try multiplying the big one with the small one, and the two middle ones.
$$p(p+1)(p+2)(p+3) + 1 = (p^2 + 3p)(p^2 + 3p + 2) + 1$$
Now those terms are nearly the same. How can we force them together? I'm going to use the basic but sometimes-overlooked fact that $xy = (x+1)y - y$, and likewise $x(y + 1) = xy + x$.
$$\begin{align*}
(p^2 + 3p)(p^2 + 3p + 2) + 1 &= (p^2 + 3p + 1)(p^2 + 3p + 2) - (p^2 + 3p + 2) + 1 \\
&= (p^2 + 3p + 1)(p^2 + 3p + 1) + (p^2 + 3p + 1) - (p^2 + 3p + 2) + 1 \\
&= (p^2 + 3p + 1)^2
\end{align*}$$
Tada.
Best Answer
First, note that $n^2$ can't be one of the elements as the other element would also need to be a perfect square for the product to be a perfect square. As gnasher729 commented to the answer, this is why $\left(n + 1\right)^2$ is not included.
Assume there are $2$ such elements, $n^2 + a$ and $n^2 + b$, with $a \neq b$, $1 \le a, b \le 2n$ and, WLOG, $a \lt b$. Thus, consider their product to be a perfect square of $n^2 + c$, i.e.,
$$\left(n^2 + a\right)\left(n^2 + b\right) = \left(n^2 + c\right)^2 \tag{1}\label{eq1}$$
for some integer $a \lt c \lt b$. As such, for some positive integers $d$ and $e$, we have that
$$a = c - d \tag{2}\label{eq2}$$ $$b = c + e \tag{3}\label{eq3}$$
Substitute \eqref{eq2} and \eqref{eq3} into \eqref{eq1} to get
$$\left(n^2 + \left(c - d\right)\right)\left(n^2 + \left(c + e\right)\right) = \left(n^2 + c\right)^2 \tag{4}\label{eq4}$$
Expanding both sides gives
$$n^4 + 2cn^2 + \left(e - d\right)n^2 + c^2 + c\left(e - d\right) - ed = n^4 + 2cn^2 + c^2 \tag{5}\label{eq5}$$
Removing the common terms on both sides and moving the remaining terms, apart from the $n^2$ one, to the right gives
$$\left(e - d\right)n^2 = -c\left(e - d\right) + ed \tag{6}\label{eq6}$$
Note that $e \le d$ doesn't work because the LHS becomes non-positive but the RHS becomes positive. Thus, consider $e \gt d$, i.e., let
$$e = d + m, \text{ where } m \ge 1 \tag{7}\label{eq7}$$
Using \eqref{eq3} - \eqref{eq2}, this gives
$$b - a = e + d \lt 2n \Rightarrow 2d + m \lt 2n \Rightarrow d \lt n - \frac{m}{2} \tag{8}\label{eq8}$$
Also,
$$ed = \left(d + m\right)d \lt \left(n + \frac{m}{2}\right)\left(n - \frac{m}{2}\right) = n^2 - \frac{m^2}{4} \lt n^2 \tag{9}\label{eq9}$$
Since $e \gt d$ means that $-c\left(e - d\right) \lt 0$, the RHS of \eqref{eq6} cannot be a positive integral multiple of $n^2$, so it can't be equal to the LHS.