Converse to "If $F$ algebraically closed, then all matrices over $F$ are trigonalisable" The following Corollary is from Hoffman and Kunze's book, Linear Algebra:

Corollary. Let $\mathbb{F}$ be an algebraically closed field, e.g.
  the complex number field. Every $n\times n$ matrix over $\mathbb{F}$
  is similar over $\mathbb{F}$ to a triangular matrix.

Is there a converse to this Corollary when $n\geq 2$ ? I.e. If every $n\times n$ matrix over $\mathbb{F}$ is similar over $\mathbb{F}$ to a triangular matrix, then $\mathbb{F}$ is algebraically closed.
 A: If $\mathbb{F}$ is not algebraically closed, let $p(x)$ be a monic irreducible polynomial of degree greater than 1, and let $M$ be its companion matrix. Then $M$ does not have any eigenvectors (since it has no eigenvalues, as its characteristic polynomial is $p(x)$), so it cannot be similar to an upper triangular matrix.

As Gerry Myerson notes in a comment, a different interpretation of your question is:

Suppose that there exists $n\geq 2$ such that every $n\times n$ matrix with coefficients in $\mathbb{F}$ is similar to an upper triangular matrix; is $\mathbb{F}$ algebraically closed?

In this case, the answer is "no". Let $\mathbb{F}$ be the field of all complex numbers that are constructible using straightedge and compass; that is, all numbers that lie in an extension whose degree is a power of $2$. Over this field, every quadratic polynomial splits. Therefore, every $2\times 2$ matrix has a Jordan canonical form, which is of course upper triangular. So every $2\times 2$ matrix over $\mathbb{F}$ is similar to an upper triangular matrix, but $\mathbb{F}$ is not algebraically closed, since it famously does not contain $\sqrt[3]{2}$.
Added.
What we have is:
Proposition. For a fixed $n$, every $n\times n$ matrix with coefficients in $\mathbb{F}$ is similar to an upper triangular matrix if and only if every polynomial of degree at most $n$ with coefficients in $\mathbb{F}$ splits over $\mathbb{F}$.
Proof. If every polynomial of degree at most $n$ splits over $\mathbb{F}$, then the characteristic polynomial of $M$ splits, hence $M$ has a Jordan canonical form, which is in upper triangular form.
Conversely, if $f(x)$ is a polynomial of degree at most $n$, let $p(x) = x^{n-m}f(x)$, and let $M$ be the companion matrix of $p(x)$. The characteristic polynomial of $M$ is $x^{n-m}f(x)$; we know that $M$ is similar to a triangular matrix $T$. But the characteristic polynomial of an upper triangular matrix is a product of linear factors, and since the characteristic polynomial is a similarity invariant, the characteristic polynomial of $M$, i.e. $x^{n-m}f(x)$, splits into linear factors over $\mathbb{F}$. Hence, $f(x)$ splits into linear factors over $\mathbb{F}$, as claimed. $\Box$
