What is the maximum number of pieces that a pizza can be cut into by 7 knife cuts? (NBHM 2005)

This is a known problem: it is appropriately called the Circle Cutting Problem.

I suggest to you to read the linked page --- it is quite informative. However, the important part is that with $n$ cuts, you can divide a circle into at most $f(n)$ parts, where:

$$f(n) = \frac12\left(n^2 + n + 2\right)$$

In your case, $f(7) = \frac12(49+7+2) = \frac12(58) = 29$.

Let's apply Euler's polyhedron formula$$1-V+E-F+1=0$$ to the plane graph formed by the $n$ chords (straight knife cuts), assumed to be pairwise intersecting with no three concurrent, and the rim of the pizza. The number we want is $$f(n)=F-1=E-V+1$$ since we don't want to count the outer face which has no pizza. The number of vertices is$$V=\binom n2+2n,$$i.e., $\binom n2$ points where two chords cross, and $2n$ points where chords meet the rim. The number of edges is$$E=n^2+2n,$$i.e., each of the $n$ chords is cut into $n$ segments, and the rim in cut into $2n$ arcs. Our final answer is $$f(n)=E-V+1=n^2+2n-\binom n2-2n+1=\frac{n^2+n+2}2=\binom{n+1}2+1$$and so, for $7$ cuts,$$f(7)=\binom82+1=29.$$

When you make a straight cut through a whole or already-sliced pizza, you divide each piece that you slice through into two pieces; that is, you create $p$ new pieces, where $p$ is the number of pieces you slice through. Your cut starts in some piece, and then crosses into a new piece for each line it traverses; so $p=c+1$, where $c$ is the number of cuts you slice through. And for the $n$-th cut, you can slice through at most all of the preceding $n-1$ cuts, so $p=c+1\le n$.

Putting this together, the maximum number of pieces after $n$ cuts is at most $$ 1+1+2+3+\ldots+n=1+n(n+1)/2, $$ or $29$ pieces after $7$ cuts.