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3 added 180 characters in body

Here's a sketch proof of 2, sort of in the same spirit as Jeff Strom's answer:

These statements have equivalent formulations involving strictly commutative squares.

Denote a typical square by $\mathcal X$, with last space $X$ and two spaces $X_1$ and $X_2$ mapping into $X$, and $X_{12}$ mapping into both. It's called a homotopy pushout square if the resulting map from the homotopy pushout of $X_1\leftarrow X_{12}\to X_2$ to $X$ is a weak equivalence, and likewise for pullback.

Use the fact that if $X$ is the union of open sets $X_1$ and $X_2$ with intersection $X_{12}$ then the resulting square is a homotopy pushout.

Also use this converse: Any homotopy pushout square admits an equivalence from such an "open triad square"--a map that is a weak equivalence in all four corners.

Now let $\mathcal X\to \mathcal Y$ be a cube, a map of squares. Suppose that $\mathcal Y$ is a homotopy pushout square and that the four side squares are homotopy pullback squares.

Wlog the map $X\to Y$ is a fibration and the side squares are pullback (not just homotopy pullback) squares. Make an open triad square $\mathcal Y'$ and an equivalence $\mathcal Y'\to \mathcal Y$. Pull back to get a new square $\mathcal X'$. This is now also of the open triad kind, so it's a homotopy pushout square. And its map to $\mathcal X$ is an equivalence, so the latter is also a homotopy pushout square.

The other theorem, 1, implies 2 (for Top) but I don't think you can go the other way. And when I try to prove 1 for Top I end up using quasifibrations.

EDIT: The first theorem really seems deeper than the second. In the category of sets (with equivalence=iso, holim=lim, hocolim=colim) the second is true and the first is false.

2 added 18 characters in body

Here's a sketch proof of 2, sort of in the same spirit as Jeff Strom's answer:

These statements have equivalent formulations involving strictly commutative squares.

Denote a typical square by $\mathcal X$, with last space $X$ and two spaces $X_1$ and $X_2$ mapping into $X$, and $X_{12}$ mapping into both. It's called a homotopy pushout square if the resulting map from the homotopy pushout of $X_1\leftarrow X_{12}\to X_2$ to $X$ is a weak equivalence, and likewise for pullback.

Use the fact that if $X$ is the union of open sets $X_1$ and $X_2$ with intersection $X_{12}$ then the resulting square is a homotopy pushout.

Also use this converse: Any homotopy pushout square admits an equivalence from such an "open triad square"--a map that is a weak equivalence in all four corners.

Now let $\mathcal X\to \mathcal Y$ be a cube, a map of squares. Suppose that $\mathcal Y$ is a homotopy pushout square and that the four side squares are homotopy pullback squares.

Wlog the map $X\to Y$ is a fibration and the side squares are pullback (not just homotopy pullback) squares. Make an open triad square $\mathcal Y'$ and an equivalence $\mathcal Y'\to \mathcal Y$. Pull back to get a new square $\mathcal X'$. This is now also of the open triad kind, so it's a homotopy pushout square. And its map to $\mathcal X$ is an equivalence, so the latter is also a homotopy pushout square.

The other theorem, 1, implies 2 (for Top) but I don't think you can go the other way. And when I try to prove 1 for Top I end up using quasifibrations.

1

Here's a sketch proof of 2, sort of in the same spirit as Jeff Strom's answer:

These statements have equivalent formulations involving strictly commutative squares.

Denote a typical square by $\mathcal X$, with last space $X$ and two spaces $X_1$ and $X_2$ mapping into $X$, and $X_{12}$ mapping into both. It's called a homotopy pushout square if the resulting map from the homotopy pushout of $X_1\leftarrow X_{12}\to X_2$ to $X$ is a weak equivalence, and likewise for pullback.

Use the fact that if $X$ is the union of open sets $X_1$ and $X_2$ with intersection $X_{12}$ then the resulting square is a homotopy pushout.

Also use this converse: Any homotopy pushout square admits an equivalence from such an "open triad square"--a map that is a weak equivalence in all four corners.

Now let $\mathcal X\to \mathcal Y$ be a cube, a map of squares. Suppose that $\mathcal Y$ is a homotopy pushout square and that the four side squares are homotopy pullback squares.

Wlog the map $X\to Y$ is a fibration and the side squares are pullback (not just homotopy pullback) squares. Make an open triad square $\mathcal Y'$ and an equivalence $\mathcal Y'\to \mathcal Y$. Pull back to get a new square $\mathcal X'$. This is now also of the open triad kind, so it's a homotopy pushout square. And its map to $\mathcal X$ is an equivalence, so the latter is also a homotopy pushout square.

The other theorem, 1, implies 2 but I don't think you can go the other way. And when I try to prove 1 I end up using quasifibrations.