A more or less mechanical approach is to work modulo $4$. Note that for any integer $k$, $k^2\equiv 0 \pmod{4}$ or $k^2\equiv 1\pmod{4}$.
So, modulo $4$, $x^2-y^2$ can only take on the values $0$, $1$, and $-1$.
More basic, and more useful, is to suppose that $x^2-y^2=n$. Then $(x-y)(x+y)=n$. Note that for any integers $x$ and $y$, the numbers $x-y$ and $x+y$ are both even or both odd. (If we want a proof, their difference $2y$ is even.)
In neither case is $(x-y)(x+y)$ twice an odd integer. You started along these lines. Note that you were one step from the end.
Remark: The reason the second idea is more useful is that when it comes to solving $x^2-y^2=n$, we express $n$ as a product $st$ of integers of the same parity, and solve the system $x-y=s$, $x+y=t$. The solution is $x=\frac{s+t}{2}$, $y=\frac{s-t}{2}$. If $n$ is twice an odd integer, then this process breaks down, because one of $s$ and $t$ will be odd and the other even, so we do not get integers $x$ and $y$.
To expand on the other solution already given,
Proof.
Assume $m$ and $n$ exist in the positive integers and that $m$ is less than $n$. If $m$ is less than $n$, then
$$
n!=1\cdot2\cdot3\cdot\dotsb\cdot m\cdot\dotsb\cdot n,
$$
which is to say that $m$ is a factor of $n!$. So,
\begin{align}
m+n! &= m+(1\cdot2\cdot\dotsb\cdot m\cdot\dotsb\cdot n) \\
&= m\left(\frac{1\cdot 2\cdot \dotsb\cdot n}{m + 1}\right)
\end{align}
Remember that $m$ is a factor of $n!$ and so $n!/m$ is still an integer.
So since $m$ is an integer that is bounded between $1$ and $n$, it stands that whatever number you pick up to $n$ can divide $m+n!$ making it composite till the $n$th integer, but $n!$ has that $n$th integer in it so the $n$th integer is also composite which means that you can pick any integer between $1$ and $n$ inclusively and it will be composite.
Best Answer
Lemma. If $a$ and $b$ are integers such that $3a^4+3a^2+1=b^2\!$, then $a=0$ and $b=\pm 1$.
Proof. We may, without loss of generality, assume that $a$ and $b$ are non-negative. Assume, contrary to the lemma, that $a \ne 0$ and $b > 1$ are integers satisfying the equation. Evidently $b$ is odd, say $b=2c+1$ for an integer $c \ge 1$. Hence \begin{align*} 3a^4 + 3a^2+1 &= (2c+1)^2 \\ 3a^2(a^2+1) &= 4c(c+1). \end{align*} Since $4 \nmid 3(a^2+1)$, regardless of the parity of $a$, we must have $2 \mid a^2$; hence $a$ is even, say $a=2d$ for an integer $d \ge 1$. Now \begin{align*} 3d^2(4d^2+1) &= c(c+1). \end{align*} We now show that this equation has no solutions with $c,d \ge 1$. Assume to the contrary that there exist integers $p,q,r,s \ge 1$ such that \begin{align*} 3d^2 &= pq, &&& c &= pr, \\ 4d^2+1 &= rs, &&& c+1 &= qs. \end{align*} Since $\gcd(p,q) \mid \gcd(pr,qs) = \gcd(c,c+1)=1$ implies $\gcd(p,q)=1$, we may write $d=uv$ with $\gcd(u,v)=1$ and consider the two possible cases.
Case 1: $p=u^2$ and $q=3v^2$. Hence $c=pr = u^2r$, and by substitution $c+1=u^2r+1 = 3v^2s$. On the other hand, $rs-1 = 4d^2=4u^2v^2$, and adding these two relations yields \begin{align*} (rs-1)+(u^2r+1) &= 4u^2v^2 + 3v^2s \\ r(s+u^2) &= v^2(4u^2+3s). \end{align*} Since $d=uv$, the equation $rs = 4d^2+1$ forces $\gcd(v,r)=\gcd(s,u)=1$. Hence $v^2 \mid (u^2+s)$, and $\gcd(s+u^2,4u^2+3s) = \gcd(s+u^2,3(s+u^2)+u^2)=\gcd(s+u^2,u^2) = \gcd(s,u^2)=1$. Thus we conclude $r=4u^2+3s$ and $v^2=s+u^2$. Substituting now gives \begin{equation*} r = 4u^2+3s = 4u^2+3(v^2-u^2) = u^2+3v^2. \end{equation*} Multiplying, $4u^2v^2+1=rs=(u^2+3v^2)(v^2-u^2)$, which implies $u^4+1=3v^2(v^2-2u^2)$. But $3 \nmid u$ (because $3 \mid q$ and $\gcd(p,q)=1)$, so $u^4+1 \equiv 2\!\pmod{3}$, a contradiction.
Case 2: $p=3u^2$ and $q=v^2$. Now $c+1=pr+1=qs$ gives $3u^2r+1=v^2s$. We have $rs-1=4d^2=4u^2v^2$, and adding the two relations yields \begin{align*} (rs-1) + (3u^2r+1) &= 4u^2v^2 + v^2s \\ r(s+3u^2) &= v^2(4u^2+s). \end{align*} As before, considering common factors leads to the conclusion $s+3u^2=v^2$ and \begin{equation*} r=4u^2+s =4u^2+(v^2-3u^2) = u^2+v^2. \end{equation*} Multiplying, $4u^2v^2+1 = rs = (u^2+v^2)(v^2-3u^2)$, which is $3u^2(u^2+2v^2)=(v^2-1)(v^2+1)$. Since $3$ can never divide $v^2+1$, we deduce $v^2-1=3u_1^2w_1$ and $v^2+1=u_2^2w_2$ where $u=u_1u_2$ and $u^2+2v^2=w_1w_2$. As $4 \nmid (v^2+1)$, we deduce that $u_2$ is odd, $w_2 \equiv 1\text{ or }2\!\pmod{4}$, and $\gcd(u_2,w_1)=1$. Adding yields $2v^2 = (v^2-1)+(v^2+1) = 3u_1^2w_1 + u_2^2w_2$, and by substitution \begin{align*} w_1w_2 &= u^2+2v^2 \\ &= u_1^2u_2^2+ (3u_1^2w_1 + u_2^2w_2) \\ w_2(w_1-u_2^2) &= u_1^2(u_2^2+ 3w_1). \end{align*} Since $\gcd(w_1-u_2^2,u_2^2+3w_1)$ divides the sum $4w_1$, and $\gcd(w_1,u_2)=1$ and $u_2$ is odd, we deduce $w_1-u_2^2=u_1^2$. Now $w_2 = u_2^2+3w_1 = u_2^2+3(u_2^2+u_1^2) = (2u_2)^2+3u_1^2 \equiv 0\text{ or }3\!\pmod{4}$, contradicting $w_2 \equiv 1\text{ or }2\!\pmod{4}$ and proving the lemma.
Theorem. The equation $X^2 +1 = Y^3$ has only one integer solution, namely $(x,y)=(0,1)$.
Proof. Assume $x$ and $y$ are integers such that $x^2+1 = y^3$. Considering the equation modulo $3$, we quickly deduce that $$x^2 = (3z+1)^3−1 = 9z(3z^2+3z+1)$$ for some integer $z$. Since $\gcd(z,3z^2+3z+1)=1$, both $z$ and $3z^2+3z+1$ must be squares. Write $z=a^2$ for an integer $a$, and hence for some integer $b$ we have $b^2 = 3z^2+3z+1 = 3a^4+3a^2+1$. Now the lemma forces $a=0$, so that $z=a^2=0$, and $y=3z+1=1$, as claimed.