$G_1, G_2$ is graph. What does it mean that: $G_1\equiv_m G_2$ ?
And what does it mean that $G_1\equiv G_2$ ?
What about consequnces ? At this moment I wouldnt like to know underlying and complex theory. I ask only for inuitions and consequences like using $m$-isomorphism to show that there is no first-order sentence about graphs.
When it comes to common isomorphism I get it. For example two graps (structures) are isomorphic functions and relation in both structures behave in the same way.
Quantifier depth is a measure on first order sentences (in a language with quantifiers) that, roughly speaking, measures how many levels of nested quantifiers ($\forall$ and $\exists$) are present in the sentence.
For example, if $\phi$ is a quantifier-free formula, then $$ \forall x \exists y \forall z \phi $$ has quantifier depth $3$. I imagine that the full definition is given in your notes somewhere.
Given a first order theory$\newcommand{\T}{\mathbb T}$ $\T$, we say that two $\T$-strutures $A$ and $B$ are elementarily equivalent, and write $A\equiv B$, if $A$ and $B$ satisfy exactly the same sentences in $\T$.
As a weakening of this, for $n\in\mathbb N$ we write $A\equiv_n B$ if $A$ and $B$ satisfy exactly the same sentences as long as those sentences have quantifier depth at most $n$.
There might be some sentence $s$ of quantifier depth $n+1$ such that $A$ satisfies $s$ but $B$ doesn't.
Why is this a useful definition? Well, suppose that you have some property $P$ of graphs and suppose that $P$ were given by some formula $s$ in the first-order theory of graphs. Let $m$ be the quantifier depth of $s$: then we know that if $G_1$ and $G_2$ are graphs such that $G_1\equiv_m G_2$, then $G_1$ satisfies property $P$ if and only if $G_2$ satisfies property $P$ (since, by the definition of $\equiv_m$, we know that $G_1$ satisfies $s$ if and only if $G_2$ satisfies $s$).
Now suppose that we are trying to show that the property $P$ is not given by any first-order sentence. Then, by the discussion above, it suffices to show that for any $n$ there exist graphs $G_1$ and $G_2$ such that $G_1\equiv_nG_2$ but that $G_1$ satisfies property $P$ and $G_2$ does not.
Update (addressing the OP's questions): If $G_1,G_2$ are two finite graphs, then $G_1$ and $G_2$ are elementarily equivalent (in the first-order theory of graphs) if and only if they are isomorphic. To see this, let $G=(\{v_1,\dots,v_j\},E(G))$ be a finite graph: then a graph $H$ satisfies the following formula:
$$ (\exists v_1\dots v_n).\bigwedge_{1\le i<j\le n}(v_i\ne v_j) \wedge (\forall v.(v=v_1)\vee\dots\vee (v=v_n))\wedge\bigwedge_{\{v_i,v_j\}\in E(G)}v_i\sim v_j\wedge\bigwedge_{\{v_i,v_j\}\not\in E(G)}\neg(v_i\sim v_j) $$
This formula asserts the existence of $n$ distinct vertices, asserts that these are the only vertices in the graph and further asserts that the edge relations between these vertices are precisely those given by the graph $G$. It is not too difficult to see that the graphs satisfying the sentence above are precisely the graphs that are isomorphic to $G$.
(Note: it is essential that the graph $G$ is finite; indeed, two infinite graphs may be elementarily equivalent but non-isomorphic.)
The quantifier depth of the above sentence is $n+1$, where $n$ is the number of vertices in the graph $G$.
Now consider the graphs you mentioned: the path of length $2^n$ and the path of length $2^n+1$. These graphs are non-isomorphic and finite, so they are not elementarily equivalent. However, if we want to use the recipe above to distinguish them, then it will take a sentence of quantifier depth $2^n+1$.
The statement that these graphs are $n$-equivalent means that we cannot distinguish between them using sentences of quantifier depth $n$ or less. To see why that is, I recommend looking into Ehrenfeucht–Fraïssé games.