Proving that the 7-adic-integers contains a square root of 2 without using Hensel's Lemma This is Exercise 3 of Ch. II, §1 in Neukirch's book Algebraische Zahlentheorie.
Since the book has not talked about Hensel's Lemma so far, I would like to solve the exercise without it. The only useful result at this point seems to be the following:

(1.4) Theorem Let $F(x_1,\dots,x_n)$ be a polynomial with integral coefficients and $p$ be a fixed prime number. Then the congruence $F(x_1,\dots,x_n) \equiv 0 \mod p^\nu$ has a solution for arbitrary $\nu \geq 1$ if and only if the equation $F(x_1,\dots,x_n) = 0$ has a solution in $\mathbb{Z}_7$.

Let $F(x) = X^2-2$. To use this result, I can try to show that $F(x_1,\dots,x_n) \equiv 0 \mod 7^\nu$ for all $v \geq 1$. Any solution $\alpha$ of $F(x) = 0$ over $\mathbb{Z}/7^\nu \mathbb{Z}$ must have the form $\alpha = a_0 + a_1 p + \dots a_{\nu-1} p^{\nu-1}$.
For the case $\nu = 1$, I got $\alpha = a_0 = 4$.
For the case $\nu = 2$, I computed $2 \stackrel{!}{=} \alpha ^2 = a_0^2 + 14 a_0 a_1 + a_2 7^2 \equiv a_0^2 + 14 a_0 a_1 \mod 49$. Since $\mathbb{Z}_7$ is the inverse limit of $\mathbb{Z}/7^\nu \mathbb{Z}$'s, it must be $a_0 =4$ again.
So in total, this is equivalent to $ \frac{2-a_0^2}{2 a_0} = -14/8 = 7 \frac{-1}{4} = 7a_1 \mod 49$. Elementary number theory says that this is equivalent to $3^{-1} = a_1 \mod 7$ which gives $a_1 = 5$.
Question: How can I make sure that this approach works for greater $\nu$'s?  In particular, how can I make sure that when I have $\text{remaining stuff} = 7^{\nu -1} a_{\nu-1} \mod 7^\nu$ that the remaining stuff is a multiple of $7^{\nu-1}$ too?
 A: The question in your title I feel is slightly different than the question you ask at the end of your post. The way you work it out is effectively what you'd be doing for Hensel's lemma and so to ensure that the approach works for larger $\nu$ would amount to proving Hensel's lemma - which is maybe not what you want, but here effectively it boils down to your derivative $F'(x) \ne 0 \mod p$ so that when you make the reduction as you did from mod 49 to mod 7, you can always solve exactly for the coefficient in the residue field by inverting the coefficient that appears here. I'll write it out more fully, but really I'm just repeating Hensel's lemma but in the special case of your problem.
Given that we have an $\alpha$ such that $F(\alpha) \equiv 0 \mod 7^\nu$, we'd like to find $a_\nu$ such that $F(\alpha+a_\nu7^\nu) \equiv 0 \mod 7^{\nu+1}$. Expanding it out,
$$(\alpha+a_\nu7^\nu)^2-2 \equiv \alpha^2-2 + a_\nu 7^\nu *2\alpha \mod 7^{\nu+1}$$
Note we could/should have written it in a suggestive way if we were keeping an eye on Hensel:
$$F(\alpha+a_\nu7^\nu) \equiv F(\alpha) + a_\nu 7^\nu F'(\alpha) \mod 7^{\nu+1}$$
Regardless, because we already know $\alpha^2-2 \equiv 0 \mod 7^\nu$ we can safely factor out $7^\nu$
$$0 \equiv 7^\nu \left( \frac{\alpha^2-2}{7^\nu} + a_\nu 2\alpha\right) \mod 7^{\nu+1}$$
This must mean,
$$0 \equiv   \frac{\alpha^2-2}{7^\nu} + a_\nu 2\alpha  \mod 7$$
Since $\alpha \equiv a_0 \mod 7$ we can safely invert it every time,
$$a_\nu \equiv - (2\alpha)^{-1} \frac{\alpha^2-2}{7^\nu} \mod 7$$
That answers your question, although I am really just using Hensel's lemma the whole time, so it's kind of cheating in my eyes. So I'd like to give a second answer that kind of mimics a continued fraction approach. If we're looking at the 7-adic integers and want to find an expansion for $\sqrt{2}$; picking the root such that $\sqrt{2} \equiv 4 \mod 7$ then for $x \in \mathbb{Z}_7$ it will look like,
$$\sqrt{2} = 4 + 7x$$
Now squaring both sides gets us
$$2 = 16 + 2*7x + 7^2 x^2$$
Rearrange this by a bit of elementary school algebra
$$x = -1 - \tfrac{7}{2}x^2$$
Since $\frac{-1}{2} = 3+3*7+3*7^2 + \cdots$ is a simple geometric series, we see that it is just a 7-adic integer. The key thing to notice here is there is a $7$ multiplying the $x^2$ term, so by plugging $x$ into itself iteratively we are always multiplying by larger powers of $7$. So we have,
$$\sqrt{2} = 4 + 7x = 4 + 7(-1 - \tfrac{7}{2}x^2)= 4 + 7(-1 - \tfrac{7}{2}(-1 - \tfrac{7}{2}(-1 - \tfrac{7}{2}(\cdots)^2)^2)^2)$$
To prove this converges in the complete space $\mathbb{Z}_7$, we can use the Banach contraction theorem with $f(x) = -1- \tfrac{7}{2}x^2$. Pick two 7-adic integers x,y then
$$|f(x)-f(y)| = \frac{1}{7}|x+y|_7|x-y|_7 \le \frac{1}{7}|x-y|_7$$
This confirms that we have a contraction mapping and so there is a unique fixed point it converges to.
A: You're pretty close. Of course you should use induction; say you have found $a_n \in \mathbb Z$ such that $a_n^2 \equiv 2$ mod $7^{n+1}$. Then I claim there is a $t \in \mathbb Z$, unique modulo $7$, such that $(a_n+7^{n+1}t)^2 \equiv 2$ mod $7^{n+2}$. Namely, we have $2-a_n^2 = 7^{n+1} \cdot x_n$ for some $x_n \in \mathbb Z$ by induction hypothesis. On the other hand, for any $t \in \mathbb Z$, $(a_n+7^{n+1}t)^2 \equiv a_n^2+2\cdot 7^{n+1}t$ mod $7^{n+2}$, so we are forced to take (and can take) $t \equiv x_n/2$ mod $7$ (i.e. here, $4x_n$ as $4=2^{-1}$ mod $7$).
The only crucial fact used is that division by $2$ is a well-defined operation modulo $7$. Think about how that generalises to other $p$, to other square roots, and then more general roots of polynomials; you can generalise the above argument and get free induction steps once you have the start of the induction. You will discover Hensel's lemma, of which the above is a very special case.
