Prove That $n(n+1)$ Can Never Be a Square Number by Showing the Atleast One of Exponents in the Prime Power Decomposition Isn't Even

Question

Can someone show me how to prove that when $n>0$, $n(n+1)$ can never be a square number by demonstrating at least one of the exponents in the prime power decomposition is not even?

Here's what I have done so far:

Without loss of generality I can say $n$ is even and $n+1$ is odd.

The prime power decomposition of $n$ and $n+1$ going from lowest divisor to highest is $n=p^{e_1}_1 p^{e_2}_2 ... p^{e_k}_k$ and $n+1=q^{f_1}_1 q^{f_2}_2 ... q^{f_j}_j$ We know because $n$ is even $2$ must be a divisor of $n$, so $n=2^{e_1} p^{e_2}_2 p^{e_3}_3 ... p^{e_n}_n$

Because $p|m$ or $p|m+1$ but not both, $(\forall i,h \in \mathbb{Z})(p_i \neq q_h)$. This means $\mathrm{gcf}(n,n+1) = 1$ which implies $nx+(n+1)y = 1$ for some integers $x$ and $y$.

As you can see I am very lost, and all help is appreciated.

EDIT: I think I was able to patch together a complete proof from the comments. If both $n$ and $n+1$ are squares then $n+1 = (2a+1)^2$ and $n = (2b)^2$, so $n+1 = 4a^2 + 4a + 1$, which means $n = 4a^2 + 4a = 2^{e_1} p^{e_2}_2 p^{e_3}_3 ... p^{e_n}_n$ which reduces to $a(a+1) = 2^{e_1 -2} p^{e_2}_2 p^{e_3}_3 ... p^{e_n}_n$. Now $a(a+1)$ must be a square. By repeating the process, some product of all the p's that equaled a must be equal to $2^{e_1 -4} p^{e_2}_2 p^{e_3}_3 ... p^{e_n}_n$ which must be a square and so forth. Eventually all the 2's will disappear and a decimal number must be equal to an integer which is impossible.

Answer 1

Not an answer strictly speaking, but assuming that $n(n+1)$ is a square, it follows that $$ 4n(n+1) = 4n^2+4n = (2n+1)^2-1 $$ is a square, too. However, there are pretty few consecutive squares.
Much easier, don't you think?

Answer 2

Consider 2 cases: 1) $n$ is a perfect square and 2) $n$ is not a perfect square

1) Assume $n(n+1)$ is a perfect square. No prime factors of $n$ divide $n+1$ and vice versa. Then, if all the exponents in the prime decomposition of $n(n+1)$ are to be even (equivalent to "$n(n+1)$ is a perfect square"), all the exponents in the prime decomposition of $n+1$ must also be even. Thus, both $n$ and $n+1$ are perfect squares. But this is only true when $n=0$ or $1$. Contradiction.

2) No prime factors of $n$ divide $n+1$ and vice versa. Thus, since $n$ is not a perfect square, $n(n+1)$ has at least 1 odd exponent in its prime power decomposition (from $n$) and is therefore not a perfect square.

This proves what we wanted to show.