No.

The natural objects in Minkowski space are Lorentz scalars, 4-vectors, etc, and there is no room for relativistic mass there.

The definition of relativistic mass is "a different mass ($\gamma m$) so that Newtonian formula (e.g $p=mv$) work in regions where Newtonian formula don't work". Definitely sus.

The relevant Minkowski space things are:

$$ u_{\mu} = (\gamma c, \gamma \vec v) $$

(Note: $\gamma$ is here, and mass is not). Then

$$ p_{\mu} = mu_{\mu}$$

You can also write:

(note: looks like a generalization of $\vec p = m\vec v$...is that not good enough?).

$$p^2 = m^2|u|^2 = (mc)^2 $$

We're done: no relativistic mass.

You can also write:

$$ p_{\mu} = mu_{\mu} = (\gamma mc, \gamma m\vec v)$$

and then call that:

$$ p_{\mu} = (E/c, \vec p) $$

with

$$ E = \gamma mc^2 $$ $$ \vec p = \gamma m \vec v$$

and then have the urge to do:

$$ m \rightarrow \gamma m $$

so:

$$ E = mc^2 $$ $$ \vec p = m\vec v $$

at all $v$, but that has nothing to do with Minkowski space.

No.

The natural objects in Minkowski space are Lorentz scalars, 4-vectors, etc, and there is no room for relativistic mass there.

The definition of relativistic mass is "a different mass ($\gamma m$) so that Newtonian formula (e.g $p=mv$) work in regions where Newtonian formula don't work". Definitely sus.

The relevant Minkowski space things are:

$$ u_{\mu} = (\gamma c, \gamma \vec v) $$

(Note: $\gamma$ is already here, and mass is not). Then

$$ p_{\mu} = mu_{\mu}$$

(note: looks like a generalization of $\vec p = m\vec v$...is that not good enough?).

$$p^2 = m^2|u|^2 = (mc)^2 $$

We're done: no relativistic mass.

You can also write:

$$ p_{\mu} = mu_{\mu} = (\gamma mc, \gamma m\vec v)$$

and then call that:

$$ p_{\mu} = (E/c, \vec p) $$

with

$$ E = \gamma mc^2 $$ $$ \vec p = \gamma m \vec v$$

and then have the urge to do:

$$ m \rightarrow \gamma m $$

so:

$$ E = mc^2 $$ $$ \vec p = m\vec v $$

at all $v$, but that has nothing to do with Minkowski space.