Proving that $2444_b$ is not a perfect square in any base $b$

Question

Here's the question:

Prove that there is no (positive integer) base $b$ such that $2444_b$ is a perfect square

My attempt:

Suppose, if possible, $2444_b$ is a perfect square in some base $b$.

Case 1: $3\nmid b$

By Fermat's little theorem:

$\begin{align}2444_b&=2b^3+4(b^2+b+1) \\&\equiv 2b+1(1+b+1)\\ &\equiv 2 \pmod 3\end{align}$

A perfect square $\text{mod } 3$ cannot be $2$. (Contradiction)

Case 2: $3\mid b$

Clearly, $b$ must be even because:

$2\mid 2444_b\implies 4\mid 2444_b\implies 2\mid b^3$

Thus, $6\mid b$.

I can't find a contradiction in this case.

According to WolframAlpha, the only integer solutions of $2x^3+4(x^2+x+1)=y^2$ are $(0,\pm 2)$.

Answer 1

Suppose $a$ and $b$ are integers such that $$2b^3+4b^2+4b+4=a^2.\tag{1}$$ Set $x=2b+1$ and $y=2a$ to get the minimal Weierstrass form $$y^2=x^3+x^2+3x+11,$$ which is an elliptic curve with integral points $(x,y)=(-2,\pm1)$ and $(x,y)=(1,\pm4)$, see LMFDB. Only the latter pair correspond to integral solutions to $(1)$, but with $b=0$ which is not a sensible base.