This is equivalent to showing this MOD equation has infinitely many solutions for any given positive integer $k$: $$m^2 + 7 \equiv 0\pmod{2^k}$$

When $k=1, 2, 3$, any integer $m\equiv 1\pmod{2^k}$ will satisfy the above MOD relation.

Assume that for $k\ge 3$, there exists $m$ such that $m^2+7\pmod{0}\pmod{2^k}$. Clearly, $m$ is odd.

Then, for the case of $(k+1)$, let's consider the number $\left(m+a\cdot 2^{k-1}\right)$ where $a$ is a positive integer: $$\left(m+a\cdot 2^{k-1}\right)^2 + 7\equiv m^2 + am\cdot 2^k + 7\equiv 2^k(am + b)\pmod{2^{k+1}}$$

where $b=\frac{m^2+7}{2^k}$. By the assumption of the case $k$, we know $b$ is an integer. Now, because $m$ is odd, as long as $a$ and $b$ have the same parity, $(am+b)$ will be even. This means $$2^k(am+b)\equiv p\pmod{2^{k+1}}\implies \left(m+a\cdot 2^{k-1}\right)^2 + 7\equiv 0\pmod{2^{k+1}}$$

Hence, by the principle of mathematical induction, we claim $m^2+7\equiv 0\pmod{2^k}$ is always solvable. This means there exist infinitely many square numbers in the form of $(n\cdot 2^k + 7)$.