Prove that product of the first $n$ prime numbers plus $1$ is never a perfect square

Question

Let $$m = 2 \cdot 3 \cdot\cdots\cdot p_n$$ be the product of the fist $n$ primes.
How can we prove that $m+1$ is not a perfect square ?
My attempt was maybe to proof by contradiction that for example:
$m+1=x^2$
$m=x^2-1$
$m=(x+1)(x-1)$
But then I got stuck, how to continue ?

Answer 1

As hinted in one of the comments, observe that an odd perfect square is always in the form of $1 + 4n$ for some $n$. So if $m+1$ is a perfect square then $m + 1 = 1 + 4n$ for some $n$ which implies that $4$ divides $m$ which is not possible.