I'll use "complex" to mean "nonnegatively graded (cochain) complex". Of course if you have a bounded below cochain complex, up to a shift one may assume we are in this situation.
Here's a direct argument, that singles out the essential bits of the "model-category argument" that I sketched in the comments. It turns out that the important things you need are as follows (and they're not model-category-theoretic in nature, you actually need them to prove that the injective model-structure exists, or at least the first 2) :
1) Suppose $f:A\to B$ is a degree-wise monomorphism that is also a quasi-isomorphism, and $g:C\to D$ is a degree-wise epimorphism with a kernel that is a complex of injectives, then $f$ has the left lifting property against $g$, that is, every commutative square
$$\require{AMScd} \begin{CD}
A @>{}>> C\\ @V{f}VV @VV{g}V\\
B @>>{}> D
\end{CD}$$
has a diagonal map $B\to C$ that makes the whole thing commute (that is, the two triangles commute)
2) If $\alpha\circ \beta$ and $\beta$ are quasi-isomorphisms, then so is $\alpha$
3) Let $I\to J$ be a quasi-isomorphism of complexes of injectives. Then it is a chain homotopy equivalence.
With these 3 tools, you can conclude in the following way :
Note that an injective resolution $A\to I$ is a degree-wise monomorphism that is also a quasi-isomorphism, and that if $J$ is a complex of injectives, $J\to 0$ is a degree-wise epimorphism with degree-wise injective kernel. Therefore, if $A\to I,A\to J$ are injective resolutions, we have the following square
$$\require{AMScd} \begin{CD}
A@>{}>> I\\ @V{}VV @VV{}V\\
J @>>{}> 0
\end{CD}$$
that satisfies the hypotheses of 1), so that there is a lift $J\to I$. The top triangle consists of 2 quasi-isomorphisms and the map $J\to I$ so by 2) $J\to I$ is a quasi-isomorphism. Thus by 3) it is a chain homotopy equivalence.
Proving 2) is fairly easy : prove that it holds for isomorphisms in any category, then note that a quasi-isomorphism is just some arrow $\alpha$ with $H^*(\alpha)$ being an isomorphism.
The idea for 1) is that each $C^n\to D^n$ has injective cokernel, so locally it is $C^n\simeq D^n\oplus I^n$ for some $I^n$, which you can then use to get a lift $B\to C$ by induction, because you already have $B^n\to D^n$, and you just need to pick a lift $B^n\to I^n$ of $A^n\to I^n$, but you can because $I^n$ is injective. Then you need to adjust the lift a bit to fit with the differentials that came before. For that you can use the cokernel of $A\to B$, which, by the long exact sequence in cohomology, is acyclic; and its acyclicity will allow you to do the adjustments. If you want I can give more details.
To prove 3), following Weibel, we first use a lemma : let $C$ be an exact cochain complex and $I$ a complex of injectives. Then any map $C\to I$ is nullhomotopic. This is a pretty standard/straightforward exercise in homological algebra that I'll leave to you.
Then let $f:I\to J$ denote our quasi-isomorphism, and consider $cone(f)$ : since $f$ is a quasi-isomorphism, it is exact. Moreover, we have a natural map $cone(f) \to TI$ where $TI^n = I^{n+1}$ which is a complex of injectives. It follows that the map $cone(f) \to TI$ is nullhomotopic.
Then one checks by some computation (see Weibel, proof of 10.4.6) that this provides a map $J\to I$ that, composed with $I\to J$ is homotopic to $id_I$.
In particular, the composition $I\to J\to I$ is a quasi-isomorphism, so the map $J\to I$ must be one too. Therefore, by what we just did, we get a map $I\to J$ such that the composite $J\to I\to J$ is homotopic to $id_J$.
Then in the homotopy category, $J\to I$ has a right-inverse and a left-inverse, it is therefore an isomorphism, and the two inverses are equal, so by going back to the original category, we see that the map $J\to I$ is a quasi-inverse to our original $I\to J$ (since its other inverse is homotopic to it, and homotopies compose)
We used that our complexes were nonnegatively graded in the proofs of 1) and 3) : in 1), it's to start off the induction and define the lift; in 3) it's to claim that a chain map from an exact complex to a complex of injectives is nullhomotopic, again it's to start off the induction (I left that as an exercise for you !)
ADDED (more words on 1) ) : I strongly recommend that you draw the diagrams yourself ! I will try to draw some here, but MSE is really annoying with commutative diagrams so I won't draw that many, and they won't be that good.
Let $I$ denote the kernel of $C\to D$: it is a complex of injectives, so locally we have a splitting $C_n\simeq D_n\oplus I_n$. Of course, a priori, this splitting is not a splitting of complexes, this is why we need to work to adjust the lifts along the way.
Let $K$ denote the cokernel of $A\to B$. By hypothesis and by the long exact sequence in cohomology, it is an exact complex. Note that the hypothesis of quasi-isomorphism together with the fact that our complexes are bounded and that $A\to B$ is a pointwise monomorphism in positive degree imply that $A^0\to B^0$ is a monomorphism too.
Let us build the lift by induction (we will call it $h$)
Let $k^0 : B^0 \to I^0$ be a lift of $A^0\to C^0\to I^0$ to $B^0$ : it exists as $I^0$ is injective and $A^0\to B^0$ is a monomorphism. Then $h^0: B^0\to D^0\oplus I^0 \overset{\simeq}\to C^0$ is clearly a lift on the degree $0$.
Now let's assume we have compatible lifts up to degree $n$ :
we have $h^k:B^k\to C^k$ that makes the square with diagonal commute in degree $k$ ,and $d^k_C\circ h^k = h^{k+1}\circ d_B^k$ for $k+1\leq n$.
Our goal is to build $h^{n+1}$. Note that $C^{n+1}\simeq I^{n+1}\oplus D^{n+1}$ and the map $B^{n+1}\to D^{n+1}$ is fully determined so we only need to get $B^{n+1}\to I^{n+1}$. Now $A^{n+1}\to B^{n+1}$ is a monomorphism and $I^{n+1}$ is injective, so we can extend it to $B^{n+1}$ : we get $\tilde{h}^{n+1} : B^{n+1}\to C^{n+1}$ that makes the square in degree $k+1$ commute, the problem being that it isn't necessarily compatible with the differentials: the discrepancy is "measured" by $a:= d_C^n\circ h^n - \tilde{h}^{n+1}\circ d_B^n$. (here I recommend drawing a sort of cubical diagram which consists in the commutative square in degrees $n,n+1$, the differentials between the two, and $h^n,\tilde{h}^{n+1}$ to see what's happening)
If we can write $a=\epsilon\circ d_B^n$ with $\epsilon$ with values in $I^{n+1}$, by putting $h^{n+1} = \tilde{h}^{n+1}+\epsilon$, we will be done and the induction may proceed (check this !). To do so we will use injectivity of $I^{n+1}$ and exactness of $K$.
Note that $a$ lands in $I^{n+1}$ and vanishes on $A^n$ (this is a simple diagram-chase by remembering how $\tilde{h}^{n+1}$ was defined and remembering why a few diagrams commute -use the diagram you drew above and compose with the correct things), therefore it factors as $B^n\to K^n\overset{\alpha}\to I^{n+1}\to C^{n+1}$
Then, note that by the induction hypothesis (compatibility of $h^n$ with lower degree differentials), $a\circ d_B = 0$ and this forces $\alpha\circ d_K=0$ (as the projection $B\to K$ is an epimorphism), so $\alpha$ factors as $K^n\to K^n/\mathrm{im}(d) \to I^{n+1}$. But $K$ is exact, therefore $K^n/\mathrm{im}(d) = K^n/\ker(d) = \mathrm{im}(d)$, and so we have a map $\mathrm{im}(d) \overset{\beta}\to I^{n+1}$
$\mathrm{im}(d) \hookrightarrow K^{n+1}$ and the injectivity of $I^{n+1}$ allow us to extend $\beta$ to $b: K^{n+1}\to I^{n+1}$, and thus by composing with the projection $B^{n+1}\overset{\epsilon}\to I^{n+1}$. It now remains to check that this $\epsilon$ satisfies what we wanted. First of all, it lands in $I^{n+1}$, as desired. And then the equation with $\epsilon\circ d$ is just a consequence of how everything was defined, and I'll leave that verification to you (it involves drawing a diagram containing $B^n, B^{n+1}, K^n, K^{n+1}, I^{n+1}, \mathrm{im}(d),C^{n+1}$)
Then proceed in the induction, and at the end you get your chain morphism $h:B\to C$ filling in the diagram. Here, we used that everything was bounded below (at $0$, but of course this is a notational simplification) to start off the induction.
