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$\newcommand\BdR{B_\text{dR}^+}\newcommand\Ainf{A_\text{inf}}$Thank you for the question! Actually, you've caught me out. Though I didn't realize it at the time, I was indeed cheating and should have said things more carefully. Hopefully I can partially atone here. Below I will describe two reasons why I was cheating, one mathematical and one moral. Edit: In his answer, Z.M explains that the mathematical reason does not apply! So I was only morally cheating :)

But for now let me just give the take-away:

I should not have implied that we give a new construction of $\BdR$. Rather I should have said that we give a universal property for $\BdR$: it is the universal pro-nilpotent thickening in solid rings.

The two reasons why I was cheating:

1. I don't think it's actually true that the contangent complex being $F[1]$ implies, for purely formal reasons, a full universal one-parameter formal deformation. Certainly it does give the first-order deformation. But that only uses that the $H_1$ is $F$ and the $H_0$ is $0$, not that the others vanish. Using the vanishing of the cotangent complex in higher degrees, you do get a good deal of knowledge about higher deformations by iteratively using transitivity triangles, and it feels like it should be saying a lot, but as far as I know it's still incomplete. To get around this problem and actually make a full one-parameter formal deformation, one option is to work relative to an already-existing one. So in the perfect field case, you can use that $\mathbb Z$ is already a one-parameter deformation of $\mathbb F_p$ and that the cotangent complex of $k$ over $\mathbb F_p$ vanishes; this is the approach discussed in section six of Bhargav's notes, which is where I learned it. In the perfectoid case, you at least see in the same way that if you construct $\BdR$ by hand for a base perfectoid, then you get it for all perfectoids over that one. Not as snazzy as what I claimed, but it's also not nothing: for example, $\Ainf$ and $\BdR$ for the $p$-cyclotomic extension are very explicit.

2. Even if it were true that you get the construction for formal reasons from the cotangent calculation, the way I know how to get the cotangent calculation uses $\Ainf$. So it would still be morally, if not mathematically, incorrect to claim this is a new construction of $\BdR$.

$\newcommand\BdR{B_\text{dR}^+}\newcommand\Ainf{A_\text{inf}}$Thank you for the question! Actually, you've caught me out. Though I didn't realize it at the time, I was indeed cheating and should have said things more carefully. Hopefully I can partially atone here. Below I will describe two reasons why I was cheating, one mathematical and one moral.

Edit: In his answer, Z.M explains that the mathematical reason does not apply, and in Peter's answer he explains that the moral reason does not apply! So I was overly pessimistic on both counts, and in spite my ignorance my claims from my talk are perfectly substantiated!

But for now let me just give the take-away:

I should not have implied that we give a new construction of $\BdR$. Rather I should have said that we give a universal property for $\BdR$: it is the universal pro-nilpotent thickening in solid rings.

The two reasons why I was cheating:

1. I don't think it's actually true that the contangent complex being $F[1]$ implies, for purely formal reasons, a full universal one-parameter formal deformation. Certainly it does give the first-order deformation. But that only uses that the $H_1$ is $F$ and the $H_0$ is $0$, not that the others vanish. Using the vanishing of the cotangent complex in higher degrees, you do get a good deal of knowledge about higher deformations by iteratively using transitivity triangles, and it feels like it should be saying a lot, but as far as I know it's still incomplete. To get around this problem and actually make a full one-parameter formal deformation, one option is to work relative to an already-existing one. So in the perfect field case, you can use that $\mathbb Z$ is already a one-parameter deformation of $\mathbb F_p$ and that the cotangent complex of $k$ over $\mathbb F_p$ vanishes; this is the approach discussed in section six of Bhargav's notes, which is where I learned it. In the perfectoid case, you at least see in the same way that if you construct $\BdR$ by hand for a base perfectoid, then you get it for all perfectoids over that one. Not as snazzy as what I claimed, but it's also not nothing: for example, $\Ainf$ and $\BdR$ for the $p$-cyclotomic extension are very explicit.

2. Even if it were true that you get the construction for formal reasons from the cotangent calculation, the way I know how to get the cotangent calculation uses $\Ainf$. So it would still be morally, if not mathematically, incorrect to claim this is a new construction of $\BdR$.

$\newcommand\BdR{B_\text{dR}^+}\newcommand\Ainf{A_\text{inf}}$Thank you for the question! Actually, you've caught me out. Though I didn't realize it at the time, I was indeed cheating and should have said things more carefully. Hopefully I can partially atone here. Below I will describe two reasons why I was cheating, one mathematical and one moral. But for now let me just give the take-away:

I should not have implied that we give a new construction of $\BdR$. Rather I should have said that we give a universal property for $\BdR$: it is the universal pro-nilpotent thickening in solid rings.

The two reasons why I was cheating:

1. I don't think it's actually true that the contangent complex being $F[1]$ implies, for purely formal reasons, a full universal one-parameter formal deformation. Certainly it does give the first-order deformation. But that only uses that the $H_1$ is $F$ and the $H_0$ is $0$, not that the others vanish. Using the vanishing of the cotangent complex in higher degrees, you do get a good deal of knowledge about higher deformations by iteratively using transitivity triangles, and it feels like it should be saying a lot, but as far as I know it's still incomplete. To get around this problem and actually make a full one-parameter formal deformation, one option is to work relative to an already-existing one. So in the perfect field case, you can use that $\mathbb Z$ is already a one-parameter deformation of $\mathbb F_p$ and that the cotangent complex of $k$ over $\mathbb F_p$ vanishes; this is the approach discussed in section six of Bhargav's notes, which is where I learned it. In the perfectoid case, you at least see in the same way that if you construct $\BdR$ by hand for a base perfectoid, then you get it for all perfectoids over that one. Not as snazzy as what I claimed, but it's also not nothing: for example, $\Ainf$ and $\BdR$ for the $p$-cyclotomic extension are very explicit.

2. Even if it were true that you get the construction for formal reasons from the cotangent calculation, the way I know how to get the cotangent calculation uses $\Ainf$. So it would still be morally, if not mathematically, incorrect to claim this is a new construction of $\BdR$.

$\newcommand\BdR{B_\text{dR}^+}\newcommand\Ainf{A_\text{inf}}$Thank you for the question! Actually, you've caught me out. Though I didn't realize it at the time, I was indeed cheating and should have said things more carefully. Hopefully I can partially atone here. Below I will describe two reasons why I was cheating, one mathematical and one moral. Edit: In his answer, Z.M explains that the mathematical reason does not apply! So I was only morally cheating :)

But for now let me just give the take-away:

I should not have implied that we give a new construction of $\BdR$. Rather I should have said that we give a universal property for $\BdR$: it is the universal pro-nilpotent thickening in solid rings.

The two reasons why I was cheating:

1. I don't think it's actually true that the contangent complex being $F[1]$ implies, for purely formal reasons, a full universal one-parameter formal deformation. Certainly it does give the first-order deformation. But that only uses that the $H_1$ is $F$ and the $H_0$ is $0$, not that the others vanish. Using the vanishing of the cotangent complex in higher degrees, you do get a good deal of knowledge about higher deformations by iteratively using transitivity triangles, and it feels like it should be saying a lot, but as far as I know it's still incomplete. To get around this problem and actually make a full one-parameter formal deformation, one option is to work relative to an already-existing one. So in the perfect field case, you can use that $\mathbb Z$ is already a one-parameter deformation of $\mathbb F_p$ and that the cotangent complex of $k$ over $\mathbb F_p$ vanishes; this is the approach discussed in section six of Bhargav's notes, which is where I learned it. In the perfectoid case, you at least see in the same way that if you construct $\BdR$ by hand for a base perfectoid, then you get it for all perfectoids over that one. Not as snazzy as what I claimed, but it's also not nothing: for example, $\Ainf$ and $\BdR$ for the $p$-cyclotomic extension are very explicit.

2. Even if it were true that you get the construction for formal reasons from the cotangent calculation, the way I know how to get the cotangent calculation uses $\Ainf$. So it would still be morally, if not mathematically, incorrect to claim this is a new construction of $\BdR$.

$\newcommand\BdR{B_\text{dR}}\newcommand\Ainf{A_\text{inf}}$Thank you for the question! Actually, you've caught me out. Though I didn't realize it at the time, I was indeed cheating and should have said things more carefully. Hopefully I can partially atone here. Below I will describe two reasons why I was cheating, one mathematical and one moral. But for now let me just give the take-away:

I should not have implied that we give a new construction of $\BdR$. Rather I should have said that we give a universal property for $\BdR$: it is the universal pro-nilpotent thickening in solid rings.

The two reasons why I was cheating:

1. I don't think it's actually true that the contangent complex being $F[1]$ implies, for purely formal reasons, a full universal one-parameter formal deformation. Certainly it does give the first-order deformation. But that only uses that the $H_1$ is $F$ and the $H_0$ is $0$, not that the others vanish. Using the vanishing of the cotangent complex in higher degrees, you do get a good deal of knowledge about higher deformations by iteratively using transitivity triangles, and it feels like it should be saying a lot, but as far as I know it's still incomplete. To get around this problem and actually make a full one-parameter formal deformation, one option is to work relative to an already-existing one. So in the perfect field case, you can use that $\mathbb Z$ is already a one-parameter deformation of $\mathbb F_p$ and that the cotangent complex of $k$ over $\mathbb F_p$ vanishes; this is the approach discussed in section six of Bhargav's notes, which is where I learned it. In the perfectoid case, you at least see in the same way that if you construct $\BdR$ by hand for a base perfectoid, then you get it for all perfectoids over that one. Not as snazzy as what I claimed, but it's also not nothing: for example, $\Ainf$ and $\BdR$ for the $p$-cyclotomic extension are very explicit.

2. Even if it were true that you get the construction for formal reasons from the cotangent calculation, the way I know how to get the cotangent calculation uses $\Ainf$. So it would still be morally, if not mathematically, incorrect to claim this is a new construction of $\BdR$.

$\newcommand\BdR{B_\text{dR}^+}\newcommand\Ainf{A_\text{inf}}$Thank you for the question! Actually, you've caught me out. Though I didn't realize it at the time, I was indeed cheating and should have said things more carefully. Hopefully I can partially atone here. Below I will describe two reasons why I was cheating, one mathematical and one moral. But for now let me just give the take-away:

I should not have implied that we give a new construction of $\BdR$. Rather I should have said that we give a universal property for $\BdR$: it is the universal pro-nilpotent thickening in solid rings.

The two reasons why I was cheating:

1. I don't think it's actually true that the contangent complex being $F[1]$ implies, for purely formal reasons, a full universal one-parameter formal deformation. Certainly it does give the first-order deformation. But that only uses that the $H_1$ is $F$ and the $H_0$ is $0$, not that the others vanish. Using the vanishing of the cotangent complex in higher degrees, you do get a good deal of knowledge about higher deformations by iteratively using transitivity triangles, and it feels like it should be saying a lot, but as far as I know it's still incomplete. To get around this problem and actually make a full one-parameter formal deformation, one option is to work relative to an already-existing one. So in the perfect field case, you can use that $\mathbb Z$ is already a one-parameter deformation of $\mathbb F_p$ and that the cotangent complex of $k$ over $\mathbb F_p$ vanishes; this is the approach discussed in section six of Bhargav's notes, which is where I learned it. In the perfectoid case, you at least see in the same way that if you construct $\BdR$ by hand for a base perfectoid, then you get it for all perfectoids over that one. Not as snazzy as what I claimed, but it's also not nothing: for example, $\Ainf$ and $\BdR$ for the $p$-cyclotomic extension are very explicit.

2. Even if it were true that you get the construction for formal reasons from the cotangent calculation, the way I know how to get the cotangent calculation uses $\Ainf$. So it would still be morally, if not mathematically, incorrect to claim this is a new construction of $\BdR$.